Two resistances A and B of 10 Ω and 20 Ω respectively are connected in parallel with a 6 V battery. The total energy supplied to the circuit by the battery in 1 second will be _______.

Question

Question

This question was previously asked in

RRB Group D Previous Year Paper (Held on: 26 Aug 2022 Shift 2)

Answer 1

The correct answer is 5.4 J

Key Points

Step 1: Calculate the equivalent resistance for the parallel connection
- The formula for the equivalent resistance for resistances in parallel is: \(\frac{1}{R_eq}=\frac{1}{R_A}+\frac{1}{R_B}\)
- Given that, \(R_A = 10 Ω\) and \(R_B = 20 Ω\): \(\frac{1}{R_eq} = \frac{1}{10} + \frac{1}{20} = \frac{2}{20} + \frac{1}{20} = \frac{3}{20}\)
- So, \(R_eq = \frac{20}{3} Ω ≈ 6.67 Ω\)
Step 2: Calculate the total current using Ohm's Law
- Ohm's law states that: \(I = \frac{V}{R}\)
- Where \(V = 6 V\) (battery voltage) and\( R_eq = 6.67 Ω\): \(I = \frac{6}{6.67} ≈ 0.9 A\)
Step 3: Calculate the total power
- The power P supplied to the circuit can be calculated using the formula: \( P = V × I\)
- Substitute the values \( V = 6 V\) and \(I = 0.9 A\): \(P = 6 × 0.9 = 5.4 W\)
Step 4: Calculate the energy supplied in 1 second
- Energy is given by the formula: E = P × t, Where t = 1 second
- \(E = 5.4 × 1 = 5.4 J\)
The total energy supplied to the circuit by the battery in 1 second is 5.4 J.

Additional Information

Energy in an electrical circuit is given by the product of power and time.
Power is the rate at which energy is transferred or converted and is given by the product of voltage and current.
When resistances are connected in parallel, the total resistance is always less than the smallest individual resistance.
In a parallel circuit, the voltage across each resistance is the same.