Numerical Problems: Current Electricity | Physics Class 12 - NEET PDF Download

Q4. A wire has resistance 50 Ω at 0°C. Its temperature coefficient is 0.004 /°C. What is the resistance at 100°C?
​Sol: R = R₀(1 + α ΔT) = 50 (1 + 0.004 × 100) = 50 (1.4) = 70 Ω

Q5.  Two identical cells, each of emf 1.5 V and internal resistance 1 Ω, are connected in parallel. Find the current through an external resistance of 1.5 Ω.
Sol: First, we calculate the equivalent internal resistance of the parallel combination:
r_eq = 1/2 = 0.5 Ω
Next, we find the total resistance in the circuit:
R_total = External resistance + r_eq = 1.5 + 0.5 = 2 Ω
Finally, we calculate the current using Ohm's Law:
I = 1.5 V / 2 Ω = 0.75 A

Q6. A battery of emf 12 V and internal resistance 2 Ω is connected to a 10 Ω resistor. Find the heat produced in 5 minutes.
Sol: First, we calculate the total resistance in the circuit:
R_total = Internal resistance + External resistance = 2 Ω + 10 Ω = 12 Ω
Next, we determine the current flowing through the circuit:
I = V / R_total = 12 V / 12 Ω = 1 A
Now we calculate the heat produced in the external resistor using Joule's law of heating:
H = I² × R × t
Where:

• I = 1 A (current)
• R = 10 Ω (external resistance)
• t = 5 minutes = 300 seconds

Substituting the values:
H = 1² × 10 × 300 = 3000 J
In a Wheatstone bridge, the four resistors are 5Ω, 10Ω, 6Ω, and 12Ω. A 3V battery is connected between the opposite corners and a galvanometer between the remaining corners. Calculate the current through the galvanometer.

Q7. In a Wheatstone bridge, the four resistors are 5Ω, 10Ω, 6Ω, and 12Ω. A 3V battery is connected between the opposite corners and a galvanometer between the remaining corners. Calculate the current through the galvanometer.
Sol: The Wheatstone bridge is balanced when:
R₁/R₂ = R₃/R₄
Given resistors:

• R₁ = 5Ω
• R₂ = 10Ω
• R₃ = 6Ω
• R₄ = 12Ω

Checking the ratio:
5/10 = 0.5 and 6/12 = 0.5
Since 0.5 = 0.5, the bridge is balanced.
In a balanced Wheatstone bridge:
V_G = 0 (potential difference across galvanometer is zero)
Therefore, the current through the galvanometer I_G = 0

Q8. In a circuit with three resistors of 2Ω, 4Ω, and 6Ω connected in a triangular loop, a 12V battery is connected across the 2Ω resistor. Calculate the power dissipated by each resistor using Kirchhoff's laws.
Sol: Let's consider the following configuration:

• Resistor R₁ = 2Ω (connected directly to battery)
• Resistor R₂ = 4Ω
• Resistor R₃ = 6Ω

The battery (12V) is connected across R₁, completing the triangle with R₂ and R₃.
Kirchhoff's Voltage Law (KVL)
For the two loops in the circuit:

1. Left loop (containing battery and R₂):

   12V - I₁R₁ - I₂R₂ = 0

2. Right loop (containing R₂ and R₃):

   I₂R₂ - I₃R₃ = 0

Kirchhoff's Current Law (KCL)

At the junction point:
I₁ = I₂ + I₃
Substituting values:

1. 12 - 2I₁ - 4I₂ = 0
2. 4I₂ - 6I₃ = 0 ⇒ I₃ = (2/3)I₂
3. I₁ = I₂ + I₃ = I₂ + (2/3)I₂ = (5/3)I₂

Substitute I₁ into equation 1:
12 - 2(5/3)I₂ - 4I₂ = 0
12 = (10/3 + 4)I₂ = (22/3)I₂
I₂ = 36/22 = 18/11 ≈ 1.636 A

Now find other currents:
I₃ = (2/3)(18/11) = 12/11 ≈ 1.091 A
I₁ = (5/3)(18/11) = 30/11 ≈ 2.727 A

Power dissipated in each resistor:

• P₁ = I₁²R₁ = (30/11)² × 2 ≈ 14.876 W
• P₂ = I₂²R₂ = (18/11)² × 4 ≈ 10.710 W
• P₃ = I₃²R₃ = (12/11)² × 6 ≈ 7.140 W

Q9. A meter bridge is used to measure an unknown resistance. The known resistance is 15Ω and the balancing point is at 60 cm. If the least count in measuring the balancing point is 0.1 cm, calculate the maximum percentage error in determining the unknown resistance.
Sol: The unknown resistance (X) is given by the meter bridge formula:
X = R × (l₂/l₁)
Where:

• R = Known resistance = 15Ω
• l₁ = Length from left end to balance point = 60 cm
• l₂ = Remaining length = 100 cm - 60 cm = 40 cm
• Step 2: Calculate Unknown Resistance

X = 15Ω × (40 cm / 60 cm) = 15Ω × (2/3) = 10Ω
The percentage error in X is given by:
% error in X = % error in l₁ + % error in l₂
Since l₂ = 100 - l₁, the error in l₂ is equal to the error in l₁:
Δl₁ = Δl₂ = Least count = 0.1 cm
Percentage errors:

• % error in l₁ = (Δl₁/l₁) × 100 = (0.1/60) × 100 ≈ 0.1667%
• % error in l₂ = (Δl₂/l₂) × 100 = (0.1/40) × 100 = 0.25%

Maximum % error in X = % error in l₁ + % error in l₂
= 0.1667% + 0.25% = 0.4167%

Q10. An electric heater is designed to operate at 100 V with a power output of 1000 W. When connected to a 25 V source, the power output is
Sol: Using the power formula:
P = V²/R
Therefore, R = V²/P
For the given heater:
R = (100 V)² / 1000 W = 10000 / 1000 = 10 Ω
Using the same formula with the new voltage:
P = V²/R = (25 V)² / 10 Ω = 625 / 10 = 62.5 W