{"id":2054,"date":"2020-10-19T10:50:47","date_gmt":"2020-10-19T05:50:47","guid":{"rendered":"https:\/\/murreeroad.org\/physics10\/?p=2054"},"modified":"2020-10-20T00:29:22","modified_gmt":"2020-10-19T19:29:22","slug":"physics-10th-chapter-14-numerical-problems","status":"publish","type":"post","link":"https:\/\/murreeroad.org\/physics10\/physics-10th-chapter-14-numerical-problems\/","title":{"rendered":"Physics 10th chapter 14 Numerical Problems"},"content":{"rendered":"

<\/a>Chapter 14 Current Electricity (Numerical Problems)<\/strong><\/span><\/p>\n

14.1. A current of 3 mA is flowing through a wire for 1 minute. What is the charge flowing through the wire?\u00a0<\/a><\/strong>
\n14.2. At 100,000 \u03a9, how much current flows through your body if you touch the terminals of a 12 V battery? If your skin is wet, so that your resistance is only 1000 \u03a9, how much current would you receive from the same battery?\u00a0<\/a><\/strong>
\n14.3. The resistance of a conductor wire is 10 M\u03a9. If a potential difference of 100 volts is applied across its ends, then find the value of current passing through it in mA.\u00a0\u00a0<\/a><\/strong>
\n14.4. By applying a potential difference of 10 V across a conductor, a current of 1.5 A passes through it. How much energy would be obtained from the current in 2 minutes?<\/a><\/strong>
\n14.5. Two resistances of 2 k\u03a9 and 8 k\u03a9 are joined in series, if a 10 V battery is connected across the ends of this combination, find the following quantities:<\/a><\/strong>
\n(a) The equivalent resistance of the series combination.<\/a><\/strong>
\n(b) Current passing through each of the resistances.<\/a><\/strong>
\n(c) The potential difference across each resistance.<\/a><\/strong>
\n14.6. Two resistances of 6 k and 12 k\u03a9 are connected in parallel. A 6 V battery is \u03a9 connected across its ends, find the values of the following quantities:<\/a><\/strong>
\n(a) Equivalent resistance of the parallel combination.<\/a><\/strong>
\n(b) Current passing through each of the resistances.<\/a><\/strong>
\n(c) Potential difference across each of the resistance.<\/a><\/strong>
\n14.7. An electric bulb is marked with 220 V, 100 W. Find the resistance of the filament of the bulb. If the bulb is used 5 hours daily, find the energy in kilowatt-hour consumed by the bulb in one month (30 days).\u00a0\u00a0<\/a><\/strong>
\n14.8. An incandescent light bulb with an operating resistance of 95 \u03a9 is labelled \u201c150 W.\u201d Is this bulb designed for use in a 120 V circuit or a 220 V circuit?<\/a><\/strong>
\n14.9. A house is installed with<\/a><\/strong>
\n(a) 10 bulbs of 60 W each of which are used 5 hours daily.<\/strong><\/a>
\n(b) 4 fans of 75 W each of which run 10 hours daily.<\/a><\/strong>
\n(c) One T.V. of 100 W which is used for 5 hours daily.<\/a><\/strong>
\n(d) One electric iron of 1000 W which is used for 2 hours daily.<\/a><\/strong>
\nIf the cost of one unit of electricity is Rs.4. Find the monthly expenditure of <\/strong>electricity (one month =30 days).\u00a0\u00a0<\/strong><\/a>
\n14.10. A 100 W lamp bulb and a 4 kW water heater are connected to a 250 V supply.<\/a><\/strong>
\nCalculate\u00a0 <\/strong>(a) the current which flows in each appliance\u00a0 <\/strong>(b) the resistance of each appliance when in use.\u00a0\u00a0<\/strong><\/a>
\n14.11. A resistor of resistance 5.6 \u03a9 is connected across a battery of 3.0 V by means of a wire of negligible resistance. A current of 0.5 A passes through the resistor. Calculate<\/a><\/strong>
\n(a) Power dissipated in the resistor.<\/a><\/strong>
\n(b) Total power produced by the battery.<\/a><\/strong>
\n(c) Give the reason of difference between these two quantities.<\/a><\/a><\/strong><\/p>\n

14.1. A current of 3 mA is flowing through a wire for 1 minute. What is the charge flowing through the wire?\u00a0<\/strong><\/p>\n

Solution: Given data:<\/strong><\/p>\n

Current I = m3 mA = 3.0 x 10-3<\/sup> A<\/p>\n

Time t = 1 min = 60 sec.<\/p>\n

Q = ?<\/p>\n

Formula I = V …… by definition<\/p>\n

I = Q\/t<\/p>\n

Q = It<\/p>\n

Q = 3.0 x 10-3<\/sup> x 60<\/p>\n

Q = 180 x 10-3<\/sup><\/p>\n

Q = 180mC<\/a>
\n14.2. At 100,000 \u03a9, how much current flows through your body if you touch the terminals of a 12 V battery? If your skin is wet, so that your resistance is only 1000 \u03a9, how much current would you receive from the same battery?\u00a0<\/strong><\/p>\n

Solution: Given Data:<\/strong><\/p>\n

Resistance R = 100,000\u03a9 = 1 M\u03a9<\/p>\n

Current I = ? If battery is V = 12 Volts<\/p>\n

Find current I = ? If For same battery but R = 100\u03a9<\/p>\n

Ohm’s Law\u00a0 V = IR\u00a0 …… by definition<\/p>\n

\u21d2 I = V \/ R<\/p>\n

Putting value we get<\/p>\n

\u21d2 I = 12 \/ 100<\/p>\n

\u21d2 I = 1.2 x 10-4<\/sup> A<\/p>\n

Again Ohm’s law\u00a0 V = IR\u00a0 …… by definition<\/p>\n

\u21d2 I = V \/ R<\/p>\n

Putting value we get<\/p>\n

\u21d2 I = 12 \/ 1000<\/p>\n

\u21d2 I = 1.2 x 10-2<\/sup> A<\/a><\/p>\n

14.3. The resistance of a conductor wire is 10 M\u03a9. If a potential difference of 100 volts is applied across its ends, then find the value of current passing through it in mA.\u00a0\u00a0<\/strong><\/p>\n

Solution: Give data<\/strong><\/p>\n

Resistance R = 10M\u03a9 = 1.0 x 107<\/sup> \u03a9\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u2234 1M = 106<\/sup><\/p>\n

Current I = ? If battery is V = 100 Volts<\/p>\n

By Ohm’s law\u00a0 V = IR\u00a0 …… by definition<\/p>\n

\u21d2 I = V \/ R<\/p>\n

Putting value we get I = 100 \/ 10000000<\/p>\n

I = 1.0 x 10-5<\/sup> A<\/p>\n

I = 0.01 mA<\/a><\/p>\n

14.4. By applying a potential difference of 10 V across a conductor, a current of 1.5 A passes through it. How much energy would be obtained from the current in 2 minutes?<\/strong><\/p>\n

Solution. Give Data:<\/strong><\/p>\n

Potential difference V = 10 Volts<\/p>\n

Current I 1.5 A<\/p>\n

Energy = ?<\/p>\n

IF current passes for time t = 2 mins = 120 seconds<\/p>\n

\u2234 1min = 60 seconds<\/p>\n

First we find resistance R = ?<\/p>\n

By Ohm’s law\u00a0 V = IR\u00a0 …… by definition<\/p>\n

\u21d2 R = V \/ I<\/p>\n

Putting the values we get<\/p>\n

R = 10 \/ 1.5<\/p>\n

R = 6.7\u03a9<\/p>\n

By definition W = I2<\/sup>Rt<\/p>\n

W = (1.5)2<\/sup> x 6.7 x 120<\/p>\n

W = 1809 joules<\/a>
\n14.5. Two resistances of 2 k\u03a9 and 8 k\u03a9 are joined in series, if a 10 V battery is connected across the ends of this combination, find the following quantities:<\/strong>
\n(a) The equivalent resistance of the series combination.<\/strong>
\n(b) Current passing through each of the resistances.<\/strong>
\n(c) The potential difference across each resistance.<\/strong><\/p>\n

Solution. Given Data:<\/strong><\/p>\n

Resistance R1<\/sub> = 2k\u03a9 and R2<\/sub> = 8 k\u03a9<\/p>\n

(a) Equivalent resistance Re<\/sub> = ?<\/p>\n

(b) Current through each resistor I = ?<\/p>\n

(c) Potential across each resistor V1<\/sub> = ? and V2<\/sub> = ?<\/p>\n

(a) As the resistors are connected in series, so we know Formula:<\/p>\n

Re<\/sub> = R1<\/sub> + R2<\/sub><\/p>\n

Re = 2k\u03a9 + 8 k\u03a9<\/p>\n

Re = 10 k\u03a9<\/p>\n

(b) For series current will be same through all resistors, By Ohm’s Law<\/p>\n

V = IR<\/p>\n

I = V \/ Re<\/sub><\/p>\n

Putting the values we get<\/p>\n

I = 10 \/ 10000<\/p>\n

I = 1.0 x 10-3<\/sup> A = 1mA<\/p>\n

(c) The potential difference across R1<\/sub> is V1<\/sub> = IR1<\/sub><\/p>\n

Putting the value we get<\/p>\n

V1<\/sub> = (10-3<\/sup>) x (2 x\u00a0 103<\/sup> \u03a9)<\/p>\n

V1<\/sub> = 2 Volts<\/p>\n

Similarly the potential difference across R2<\/sub> is<\/p>\n

V2<\/sub> = IR2<\/sub><\/p>\n

Putting the value we get<\/p>\n

V2<\/sub> = (10-3<\/sup>) x (8 x\u00a0 103<\/sup> \u03a9)<\/p>\n

V2<\/sub> = 8 Volts<\/a><\/p>\n

14.6. Two resistances of 6 k and 12 k\u03a9 are connected in parallel. A 6 V battery is \u03a9 connected across its ends, find the values of the following quantities:<\/strong>
\n(a) Equivalent resistance of the parallel combination.<\/strong>
\n(b) Current passing through each of the resistances.<\/strong>
\n(c) Potential difference across each of the resistance.<\/strong><\/p>\n

Solution: Given Data:<\/strong><\/p>\n

The resistance R1<\/sub> = 6 k\u03a9 and R2<\/sub> = 12 k\u03a9<\/p>\n

Potential V = 6 Volts<\/p>\n

(a) Re = ?\u00a0 \u00a0 \u00a0(b) I1<\/sub> = ? I2<\/sub> = ? I3<\/sub> = ? I = ?<\/p>\n

(a) As the resistance are in parallel combination, we use formula<\/p>\n

1\/Re<\/sub> = 1 \/ R1<\/sub> + 1 \/ R2<\/sub><\/p>\n

1\/Re<\/sub> = 1 \/ (6 x 103<\/sup>) + 1 \/ (12 x 103<\/sup> )<\/p>\n

1\/Re<\/sub> = (2 + 1) \/ (12 x 103<\/sup>)<\/p>\n

1\/Re<\/sub> = 3 \/ (12 x 103<\/sup>)<\/p>\n

Re<\/sub> = (12 x 103<\/sup>) \/ 3<\/p>\n

Re<\/sub> = 4k\u03a9<\/p>\n

Total current can be found V = IRe<\/sub><\/p>\n

I = V \/ Re<\/sub><\/p>\n

I = 6 \/ 4 x 103<\/sup><\/p>\n

I = 1.5 x 10-3<\/sup> A<\/p>\n

I = 1.5 mA<\/p>\n

Current through R1<\/sub> is I1 = V \/ R1<\/sub> = 6 \/ 6 x 103<\/sup> = 1 mA<\/p>\n

Current through R2<\/sub> is I1 = V \/ R2<\/sub> = 6 \/ 12 x 103<\/sup> = 0.5 mA<\/p>\n

(c) As the resistance are in parallel combination so all the resistances are at the same potential V = 6 Volts<\/a><\/p>\n

14.7. An electric bulb is marked with 220 V, 100 W. Find the resistance of the filament of the bulb. If the bulb is used 5 hours daily, find the energy in kilowatt-hour consumed by the bulb in one month (30 days).\u00a0\u00a0<\/strong><\/p>\n

Solution: Given data<\/strong><\/p>\n

Potential V = 220 Volts<\/p>\n

Power P = 100 Watts<\/p>\n

Resistance R = ?<\/p>\n

It is used daily for time t = 5 hours<\/p>\n

Energy = ? (kwh)\u00a0 we know that 1 kwh = 3.6 x 106<\/sup> Joules<\/p>\n

First we find current I = ?<\/p>\n

By power formula P = IV = (V\/R).V = V2<\/sup>\/R<\/p>\n

R = V2<\/sup> \/ P<\/p>\n

Putting the values we get<\/p>\n

R = (220)2<\/sup> \/ 100<\/p>\n

R = 484\u03a9<\/p>\n

By definition W = Pt<\/p>\n

\u2234 3.6 x 106<\/sup> Joules = 1 kwh<\/p>\n

Putting the value we get<\/p>\n

W = 100 x 5 x 30 x 3600<\/p>\n

W = 15 x 3600,00 Joules<\/p>\n

W = 15 kwh<\/a><\/p>\n

14.8. An incandescent light bulb with an operating resistance of 95 \u03a9 is labelled \u201c150 W.\u201d Is this bulb designed for use in a 120 V circuit or a 220 V circuit?<\/strong><\/p>\n

Solution: Given Data<\/strong><\/p>\n

Resistance R = 95\u03a9<\/p>\n

Power = P = 150 Watts<\/p>\n

Is designed for 120 Volts or 220 Volts = ?<\/p>\n

By power formula P = V2<\/sup> \/ R<\/p>\n

\u21d2 V2<\/sup> = PR<\/p>\n

Putting values we get<\/p>\n

V2<\/sup> = 150 x 95<\/p>\n

V2<\/sup> = 14250<\/p>\n

Taking under root V = 120 Volts<\/p>\n

It is designed for 120 Volts<\/a><\/p>\n

14.9. A house is installed with<\/strong>
\n(a) 10 bulbs of 60 W each of which are used 5 hours daily.<\/strong>
\n(b) 4 fans of 75 W each of which run 10 hours daily.<\/strong>
\n(c) One T.V. of 100 W which is used for 5 hours daily.<\/strong>
\n(d) One electric iron of 1000 W which is used for 2 hours daily.<\/strong><\/p>\n

If the cost of one unit of electricity is Rs.4. Find the monthly expenditure of <\/strong>electricity (one month =30 days).\u00a0\u00a0<\/strong><\/p>\n

Solution: Given data<\/strong><\/p>\n

Appliances of different power are given<\/p>\n

Let we find the total energy consumed by the appliances.<\/p>\n

Ea<\/sub> = 10 x 60 x 5 = 3000 Wh<\/p>\n

Eb<\/sub> = 4 x 75 x 10 = 3000 Wh<\/p>\n

\u2234 3.6 x 106<\/sup> Joules = 1 kwh<\/p>\n

1 KW = 1000 W<\/p>\n

Ec<\/sub> = 250 x 2 = 500 Wh<\/p>\n

Ed<\/sub> = 1000 x 2 = 2000 Wh<\/p>\n

Et<\/sub> = Ea<\/sub> + Eb<\/sub> + Ec<\/sub> + Ed<\/sub> = 3000 + 3000 + 500 + 2000<\/p>\n

Et<\/sub> = 8500 Wh<\/p>\n

Total energy consumed for 30 days Et<\/sub> = 8500 x 30 = 255000 Wh<\/p>\n

Et<\/sub> = 225 KWh<\/p>\n

We know 1 KWh is termed as 1 unit of electricity commercially<\/p>\n

Total cost = 255 units x 4 (Rupees)<\/p>\n

Total cost = 1020 Rupees<\/a><\/p>\n

14.10. A 100 W lamp bulb and a 4 kW water heater are connected to a 250 V supply.<\/strong>
\nCalculate (a) the current which flows in each appliance (b) the resistance of each appliance when in use.\u00a0\u00a0<\/strong><\/p>\n

Solution: Given Data<\/strong><\/p>\n

Power of lamp P1<\/sub> = 100 W<\/p>\n

Power of heater P2<\/sub> = 4 kW = 4000 Watts<\/p>\n

Voltage V = 250 V<\/p>\n

(a) Current in lamp I1<\/sub> = ? and Current in heater I2<\/sub> = ?<\/p>\n

Formula for power P = IV<\/p>\n

Current through lamp I1<\/sub> = P1<\/sub>\/V<\/p>\n

Putting the values we get I1<\/sub> = 100\/250<\/p>\n

I1<\/sub> = 0.4 A<\/p>\n

Current through heater I2<\/sub> = P2<\/sub>\/V<\/p>\n

I2<\/sub> = 4000\/250<\/p>\n

I2<\/sub> = 8 A<\/p>\n

For resistance we use formula V = IR<\/p>\n

Resistance through lamp R = V \/ I1<\/sub><\/p>\n

Putting values we get<\/p>\n

R = 250 \/ 0.4<\/p>\n

R = 625\u03a9<\/p>\n

Resistance through heater R = V \/ I2<\/sub><\/p>\n

Putting values we get<\/p>\n

R = 250 \/ 8<\/p>\n

R = 31.25\u03a9<\/a>
\n14.11. A resistor of resistance 5.6 \u03a9 is connected across a battery of 3.0 V by means of a wire of negligible resistance. A current of 0.5 A passes through the resistor. Calculate<\/strong>
\n(a) Power dissipated in the resistor.<\/strong>
\n(b) Total power produced by the battery.<\/strong>
\n(c) Give the reason of difference between these two quantities.<\/strong><\/p>\n

Solution: Give Data<\/strong><\/p>\n

Resistance R = 5.6\u03a9<\/p>\n

Potential difference V = 3.0 Volts<\/p>\n

Current I = 0.5 A<\/p>\n

(a) Power dissipated P = ?<\/p>\n

(b) Total power produced by battery P =?<\/p>\n

(c) Reason<\/p>\n

(a) Formula for power P = I2<\/sup>R ( It is power dissipated in resistor)<\/p>\n

Putting value we get P = (0.5)2 x 5.6<\/p>\n

P = 1.4 Watt ———–(1)<\/p>\n

P = IV ( It is power delivered by the battery)<\/p>\n

P = (0.5) x 3.0<\/p>\n

P = 1.5 Watt ————(2)<\/p>\n

From equation (1) and (2) the difference is 1.5 – 1.4 = 0.1 Watt<\/p>\n

This power is lost by the internal resistance of the battery.<\/p>\n

Goto Top<\/a><\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

Chapter 14 Current Electricity (Numerical Problems) 14.1. A current of 3 mA is flowing through a wire for 1 minute. What is the charge flowing through the wire?\u00a0 14.2. At 100,000 \u03a9, how much current flows through your body if you touch the terminals of a 12 V battery? If your skin is wet, so… Read More »Physics 10th chapter 14 Numerical Problems<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"neve_meta_sidebar":"","neve_meta_container":"","neve_meta_enable_content_width":"off","neve_meta_content_width":0,"neve_meta_title_alignment":"","neve_meta_author_avatar":"","neve_post_elements_order":"","neve_meta_disable_header":"","neve_meta_disable_footer":"","neve_meta_disable_title":""},"categories":[9],"tags":[],"_links":{"self":[{"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/posts\/2054"}],"collection":[{"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/comments?post=2054"}],"version-history":[{"count":9,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/posts\/2054\/revisions"}],"predecessor-version":[{"id":2063,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/posts\/2054\/revisions\/2063"}],"wp:attachment":[{"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/media?parent=2054"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/categories?post=2054"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/tags?post=2054"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}