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**Q.1. A heater is designed to operate with a power of 1000 watts in a 100 volt line. It is connected in a combinations with a resistance of 10 ohms and a resistance R to a 100 volts mains as shown in the figure. What should be the value of R so that the heater operates with a power of 62.5 watts.**

**Ans. **5Î©

**Solutions. **The resistance of the heater is

The power on which it operates is 62.5 W

Since the voltage drop across the heater is 25V hence voltage drop across 10Î© resistor is (100 â€“ 25) = 75V.

This current divides into two parts. Let I_{1} be the current that passes through the heater. Therefore

25 = I_{1} Ã— 10

I_{1} = 2.5 A

Thus current through R is 5A.

Applying Ohm's law across R, we get 25 = 5 Ã— R

â‡’ R = 5Î©

**Q.2. If a copper wire is stretched to make it 0.1% longer what is the percentage change in its resistance?**

**Ans. **0.2%

**Solutions.**

** **

Let the initial length of the wire be 100 cm, then the new

Let A_{i} and A_{f} be the initial and final area of cross-section.

Then

100 Ã—A_{i} = 100.1 A_{f}

_{From (i), (ii) and (iii)}

Thus the resistance increases by 0.2%.

Alternatively for small change

**Q.3. All resistances in the diagram below are in ohms. Find the effective resistance between the points A and B.**

**Ans. **2Î©

**Solution. **

**Q.4. In the diagram shown find the potential difference between the points A and B and between the points B and C in the steady state.**

**Ans. **V_{AB} = 25 V, V_{BC} = 75 V

**Solution. **Applying Kirchoff's law in loop AQBRC

âˆ´ Potential difference between AB = 150/6=25V

âˆ´ Potential difference between BC = 100 â€“ 25 = 75V

**Q.5. ****A battery of emf 2 volts and internal resistance 0.1 ohm is being charged with a current of 5 amps. In what direction will the current flow inside the battery?****What is the potential difference between the two terminal of the battery?**

**Ans. **Positive to negative terminal, 2.5 V

**Solution. **NOTE : The current will flow from the positive terminal to the negative terminal inside the battery.

During charging the potential difference V = E + Ir = 2 + 5 Ã— 0.1 = 2.5 V

**Q.6.****State ohmâ€™s law.**

**In the circuit shown in figure, a voltmeter reads 30 volts when it is connected across 400 ohm resistance. Calculate what the same voltmeter will read when it is connected across the 300 ohm resistance.**

**Ans.**22.5 V

**Solution. **Potential difference across the 400 Î© resistance = 30 V.

Therefore, potential difference across the 300 Î© resistance = 60 â€“ 30V = 30 V. Let R be the resistance of the voltmeter.

As the voltmeter is in the parallel with the 400Î© resistance, their combined resistance is

As the potential difference of 60 V is equally shared between the 300 Î© and 400 Î© resistance. R' should be equal to 300 Î©.

Thus

which gives R = 1200Î©, is the resistance of the voltmeter.

When the voltmeter is connected across the 300Î© resistance, their combined resistance is

âˆ´ Total resistance in the ciruit = 400 + 240 = 640 Î©

âˆ´ Current in the circuit is

âˆ´ Voltmeter reading = Potential difference across 240 Î© resistance

**Q.7.In the circuit shown in fig E _{1} =3 volts, E_{2} = 2 volts, E_{3} = 1 volt and R = r_{1} = r_{2} = r_{3} = 1 ohm.**

**(i) Find the potential difference between the points A and B and the currents through each branch. **

**(ii) If r _{2 }is short circuited and the point A is connected to point B, find the currents through E_{1}, E_{2} E_{3} and the resistor R**

**Ans.**(i) 2V, 1A, 0A, 1A (ii) 1A, 2A, 1A; 2A

**Solution. **

Applying Kirchoff's law in PQRUP starting from P moving clockwise

Applying Kirchoff's law in URSTU starting from U moving clockwise

NOTE : The â€“ ve sign of I3 indicates that the direction of current in branch UTSR is opposite to that assumed.

Applying Kirchoff's law in AURBA starting from A moving clockwise.

Current through R is I_{1} + I_{2} + I_{3} = 2A

**Q.8. Calculate the steady state current in the 2-ohm resistor shown in the circuit in the figure. The internal resistance of the battery is negligible and the capacitance of the condenser C is 0.2 microfarad.**

**Ans. **0.9A

**Solution. **

Total current through the battery

**Q.9. In the circuit shown in figure E, F, G, H are cells of emf 2, 1, 3 and 1 volt respectively, and their internal resistances are 2, 1, 3 and 1 ohm respectively.**

**Calculate : (i) the potential difference between B and D and**

**(ii) the potential difference across the terminals of each cells G and H**

**Ans. **

**Solution. **Let I_{2} current flow through the branch DCB

âˆ´ By Kirchoff's junction law, current in branch DB will be I_{2} â€“ I_{1} as shown in the figure.

Applying Kirchoff's law in loop BDAB

Applying Kirchoff's law in loop BCDB, we get

Solving (i) and (ii), we get I1 = 6/13 amp

and I_{2} = 5/13amp

(i) To find the p.d. between B and D, we move from B to D

[âˆ´cell is in charging mode]

**Q.10.A part of ciucuit in a steady state along with the currents flowing in the branches, the values of resistances etc., is shown in the figure. Calculate the energy stored in the capacitor C (4ÂµF)**

**Ans.** 8 Ã— 10^{â€“4} J

**Solution. **Applying Kirchoff's first law at junction M, we get the current i_{1} = 3A

Applying Kirchoff's first law at junction P, we get current i_{2} = 1A

NOTE : No current flows through capacitor at steady state.

Moving the loop along MNO to P

Energy stored in the capacitor

**Q.11. An infinite ladder network of resistances is constructed with a1 ohm and 2 ohm resistances, as shown in fig.**

**The 6 volt battery between A and B has negligible internal resistance : (i) Show that the effective resistance between A and B is 2 ohms. (ii) What is the current that passes through the 2 ohm resistance nearest to the battery ?**

**Ans.**(ii) 1.5A

**Solution. **(i) Let the effective resistance between points C and D be R then the circuit can be redrawn as shown The effective resistance between A and B is

This resistance R_{eq} can be taken as R because if we add one identical item to infinite items then the result will almost be the same.

**Q.12.In the given circuit **

**E _{1} = 3E_{2}=2E_{3} = 6 volts R_{1} = 2R_{4} = 6 ohms R_{3} = 2R_{2} = 4 ohms C = 5 Î¼ f .**

**Find the current in R3 and the energy stored in the capacitor.**

**Ans. **1.5 A, 1.44 Ã— 10^{â€“5} J

**Solution. **

Applying Kirchoff's law in ABFGA 6 â€“ (i_{1} + i_{2}) 4 = 0 ... (i)

Applying Kirchoff's law in BCDEFB i_{2} Ã— 3 â€“ 3 â€“ 2 + 2i_{2} + (i_{2} + i_{1}) 4 = 0 ... (ii)

Putting the value of 4 (i_{1} + i_{2}) = 6 in (ii)

3i_{2} â€“ 5 + 2i_{2} + 6 = 0

Substituting this value in (i), we get

Therefore current in R_{3}

= i_{1} + i_{2} = 1.7 â€“ 0.2 = 1.5 A

To find the p.d. across the capacitor

V_{E} â€“ 2 â€“ 0.2 Ã— 2 = V_{G}

âˆ´ V_{E} â€“ V_{G} = 2.4 V

or V = 24 V

âˆ´ Energy stored in capacitor = 1/2 CV^{2}

=1/2Ã— 5 Ã— 10^{â€“6} Ã— (2.4)^{2} = 1.44 Ã— 10^{â€“5} J

**Q.13. An electrical circuit is shown in Fig. Calculate the potential difference across the resistor of 400 ohm, as will be measured by the voltmeter V of resistance 400 ohm, either by applying Kirchhoffâ€™s rules or otherwise.**

**Ans.**6.67 V

**Solution.**We can redraw the circuit as.

The equivalent resistance between G and D is

âˆ´ It is a case of balanced wheatstone bridge.

The equivalent resistance across G and B is

NOTE : Since R_{GEB} = R_{GDB} the current is divided at G into two equal parts The current I/2 further divides into two equal parts at M.Therefore the potential difference across the voltmeter

**Q.14.In the circuit shown in Figure, the battery is an ideal one, with emf V. The capacitor is initially uncharged. The switch S is closed at time t = 0.**

**(a) Find the charge Q on the capacitor at time t. (b) Find the current in AB at time t. What is its limiting value as t â†’âˆž :**

**Ans. **

**Solution. **Let at any time t charge on capacitor C be Q. Let currents are as shown in fig. Since charge Q will increase with time 't' therefore i_{1} = dQ/dt

(a) Applying Kirchoff's second law in the loop MNABM

V = (i â€“ i_{1}) R + iR or V = 2iR â€“ i_{1}R ... (i)

Similarly, applying Kirchoff's second law in loop MNSTM, we have

Eliminating i from equation (1) and (2), we get

From equation (i)

âˆ´ Current through AB

**Q.15. A thin uniform wire AB of length 1m, an unknown resistance X and a resistance of 12 Î© are connected by thick conducting strips, as shown in the figure. A battery and a galvanometer (with a sliding jockey connected to it) are also available. Connections are to be made to measure the unknown resistance X using the principle of Wheatstone bridge. Answer the following questions. **

**(a) Are there positive and negative terminals on the galvanometer? (b) Copy the figure in your answer book and show the battery and the galvanometer (with jockey) connected at appropriate points.**

**Ans. **(a) No (b) 8Î©

**Solution.**(a) No. There are no positive and negative terminals on the galvanometer.

NOTE : Whenever there is no current, the pointer of the galvanometer is at zero. The pointer swings on both side of zero depending on the direction of current.

where r is the resistance per unit length.

**Q.16. How a battery is to be connected so that the shown rheostat will behave like a potential divider? Also indicate the points about which output can be taken.**

** **

**Ans. **Battery connected across A and B. Output across A and C or B and C.

**Solution. **Battery should be connected across A and B. Output can be taken across the terminals A and C or B and C.

**Q.17. ****Draw the circuit diagram to verify Ohmâ€™s Law with the help of a main resistance of 100 W and two galvanometers of resistances 106 W and 10â€“3 W an d a source of varying emf.****Show the correct positions of voltmeter and ammeter.**

**Solution. **For the experimental verification of Ohm's law, ammeter and voltmeter should be connected as shown in the figure.

A voltmeter is a high resistance galvanometer (106Î©) which is connected in parallel with the main resistance of 100Î©.

An ammeter is a low resistance galvanometer (10â€“3Î©) which is connected in parallel with the main resistance.

**Q.18. An unknown resistance X is to be deter m in ed using resistances R _{1}, R_{2} or R_{3}. Their corresponding null points are A, B and C. Find which of the above will give the most accurate reading and why?**

**Ans. **R2

**Solution. **KEY CONEPT : At all null points the wheatstone bridge will be balanced

where R is a constant r_{1} and r_{2 }are variable.

The maximum fractional error is

**Q.19. In the given circuit, the switch S is closed at time t = 0. The charge Q on the capacitor at any instant t is given by Q(t) = Q(1 â€“ eâ€“ ^{Î±t}). Find the value of Q_{0} and Î± in terms of given parameters as shown in the circuit.**

**Ans. **

**Solution.** Given Q = Q_{0}[1 â€“ eâ€“^{Î±t}] Here Q_{0} = Maximum charge and

Now the maximum charge Q_{0} = C [V_{0} ] where V_{0} = max potential difference across C

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