Electric Resistance and Ohm’s Law Numericals Class-12 Nootan ISC Physics Ch-5 Miscellaneous Questions. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Electric Resistance and Ohm’s Law Numericals Class-12 Nootan ISC Physics Ch-5 Miscellaneous Questions
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-5 | Electric Resistance and Ohm’s Law |
| Topics | Numericals on Miscellaneous Questions |
| Academic Session | 2025-2026 |
Electric Resistance and Ohm’s Law Numericals Class-12 Nootan ISC Physics Ch-5 Miscellaneous Questions.
Que-49. Three resistors 2 Ω, 4 Ω and 5 Ω are joined in parallel and connected to a battery of 20 V and negligible internal resistance. Find (a) equivalent resistance, (b) current drawn from the battery and (c) current through each resistor.
Ans-49 (a) Let equivalent resistance is R
1/R = 1/2 + 1/4 + 1/5
=> 1/R = 19/20
=> 1/R = 20/19
(b) Total current = V/R
=> 20×19/20 = 19 A
(c) Current through
2 Ω = 20/2 = 10 A
4 Ω = 20/4 = 5 A
5 Ω = 20/5 = 4 A
voltage in parallel resistor combination remians same for all resistor.
Que-50. A 150 Ω electric heater can take a maximum current of 1.1 A. What minimum resistance should be joined in series with it so that it may be operated at 220 V mains?
Ans-50 Voltage acroos heater = iR = 1.1 x 150 = 165 V
Supply voltage = 220 V
Therefore Voltage across resister = 220 – 165 = 55 V
again current in circuit = 1.1 A
Therefore resistance of resistor = 55/1.1 = 50 Ω
Que-51. An arc lamp operated at 80 V takes 10 A current. How much resistance should be connected in series with the lamp to use it at 240 V mains so that it takes the same current?
Ans-51 Rated voltage of lamp = 80 V
Current = 10 A
Applied voltage = 240 V
Voltage to be adjusted = 240 – 80 = 160 V
Therefore required resistancce
=> V/i = 160/10 = 16 Ω
Que-52. A filament lamp of 200 Ω resistance is used at 150 V supply. How much resistance should be joined in series with it to use it at 210 V supply?
Ans-52 Current needed by lamp = V/R = 150/200 = 0.75 A
voltage applied 210 V
voltage to be adjusted = 210 – 150 = 60 V
Therefore resistance needed = V/i = 60/0.75 = 80 Ω
Que-53. Each of the two bulbs takes 2 A current at 50 V supply. They are joined in series. How much resistance should be added in series with the lamps so that on connecting them with 120 V supply, a current of 2 A flows through each of them?
Ans-53 Voltage of each bulb = 50 V
Therefore voltage of two bulb = 50 x 2 = 100 V
Voltage applied = 120 V
Voltage to be adjusted = 120 – 100 = 20 V
Current = 2 A
Therefore resistance required = 20/2 = 10 Ω
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