Two resistors, A of 6 Ω and B of 12 Ω, are connected in parallel to a battery of 3 V. The total energy supplied by the battery to the circuit in 1 second is:

Correct option is B

Sol: The correct answer is (b) 2.25 J

Given:

Resistor A: $$R_1$$ = 6$$\Omega$$

​​Resistor B: $$R_2$$ = 12$$\Omega$$

​​Voltage: V = 3V

Formula used:

Formula for total resistance $$R_{total}$$ in parallel:

$$\frac{1}{R_{total}}$$ = $$\frac{1}{R_1} + \frac{1}{R_2}$$

Solution:

​$$\frac{1}{R_{total}}$$ = $$\frac{1}{6}+ \frac{1}{12}$$

$$\frac{1}{R_{total}}$$ = $$\frac{2}{12}+\frac{1}{12} = \frac{3}{12}$$

$$\frac{1}{R_{total}}$$ = $$\frac{12}{3}$$ = $$4\Omega$$

Calculate the total current-

$$I = \frac{V}{R_{total}}$$

$$I=\frac{3V}{4\Omega}$$ = 0.75A

​​​​​Power (Energy per second)-

P=V×I

P = 3V×0.75A = 2.25W

Total Energy Supplied in 1 Second-

Energy=P×time

Energy = 2.25W × 1s= 2.25J

The total energy supplied by the battery to the circuit in 1 second is 2.25 Joules.​