Integer Answer Type Questions for JEE: Current Electricity | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. Find the voltmeter in the circuit shown in the figure.

Ans. 0
Current in the circuit I = 3/3 = 1 amp.
potential drop across AB
VAB = 3 x 1 = 3 V
Hence, no potential drop across voltmeter.

Q.2. In the adjoining circuit, the battery E₁ has an E.M.F. of 12 volts and zero internal resistance while the battery E₂ has an emf of 2 volts. If the galvanometer G reads zero, then the value of the resistance X (in ohms) is

Ans. 100
As there is no current through galvanometer.
Hence VAB = 2V
12  = 500 i + 2  ⇒ i = 1/50 amp
X. 1/50 = 2 ⇒ x = 100Ω

Q.3. Two cells ε₁ and ε₂ connected in opposition to teach other as shown in figure. The cell ε₁ is of emf 9 V and internal resistance 3 Ω the cell ε₂ is of emf 7 V and internal resistance 7Ω. Find the potential difference between the points A and B.

Ans. 8.4

Potential difference across cell ε₁ is = 9 - 0.2 × 3 = 9 - 0.6 = 8.4 V
Potential difference across ε₂; V_AB = ε₂ + 0.2r₂ = 7 + 0.2 × 7
= 7 + 1.4 = 8.4 V

Q.4. A current in a wire is given by the equation, I = 2t² - 3t + 1, the charge through cross section of wire in time interval t = 3 s to t = 5 s is

Ans. 43.34 C 

Q.5. In an atom electrons revolves around the nucleus along a path of radius 0.72 Å making 9.4 × 10¹⁸ revolution per second. Find the equivalent current. (e = 1.6 × 10^–19 C)

Ans. 1.504 A
Radius of electron orbit r = 0.72 Å = 0.72 x 10^-10 m
Frequency of revolution of electron in orbit of given atom u
= 9.4 × 10¹⁸ rev/s
(where T is time period of revolution of electron in orbit)
∴ Then equivalent current is I = e/T = ev = 1.6 x 10^-19 x  9.4 1 x 10¹⁸ = 1.504 A

Q.6. The resistance of a heating element is 99 Ω at room temperature. What is the temperature of the element if the resistance is found to be 116 Ω? (Temperature coefficient of the material of the resistor is 1.7 × 10^-4°C^-1)

Ans. 1037.1°C
Here, R₀ = 99 Ω T₀ = 27°C

⇒ T = 1010.1 + T₀ = 1010.1 + 27 = 1037.1°C

Q.7. A block of metal is heated directly by dissipating power in the internal resistance of block. Because of temperature rise, the resistance increases exponentially with time and is given by R(t) = 0.5 e^2t, where t is in second. The block is connected across a 110 V source and dissipates 7644 J heat energy over a certain period of time. This period of time is ...............× 10^–1 sec.

Ans. 5
Let t be the required time. As power is  


According to problem,
U = 7644J

or e–^2t = 0.367  
or –2lne = ln 0.367 or –2t = –1 or t = 0.5 s

Q.8. Consider the circuit shown in figure. What is the current through the battery just after the switch is closed.

Ans. 2
Just after closing of switch S
i₁ and i₃ (current through inductor is zero)
∴ i = i₂ 
i₂ = 18/9
= 2 amp

Q.9. A D.C. supply of 120 V is connected to a large resistance X. A voltmeter of resistance 10 k Ω placed in series in the circuit reads 20 V. This is an unusual use of voltmeter for measuring very  high resistance. The value of X is _______ × 10 k Ω (approx).

Ans. 5

or 20 × 10^–4x + 20 = 120
= 5 × 10⁴ Ω
= 50 kΩ

Q.10. Each resistance is of 2W. Current in resistance R (R = 2Ω) is ______ + 9.75 ampere.

Ans. 9
At y according to Kirchoff's junction law

5y – 2x = 200 ______(1)
Similarly at x

At x + 100

We get y – 2x = 50 …(4)
From (1) and (4)
y = 37.5 V
So, current through R is 18.75 A.