Two resistors, each of 10. are connected in parallel. The combination. in turn. is connected in series to a third resistor of 10 and a battery of 6V. The power supplied by the battery is:

Correct option is D

For two resistors in parallel, the formula for the equivalent resistance ReqR_{eq}Req​ is:

$$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$$​​

Where:

• R₁=10 Ω
• R₂=10 Ω

$$\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5}$$​​

Thus, the equivalent resistance of the parallel resistors is:

R_eq=5 Ω

Now, the equivalent resistance of the parallel combination (5 Ω) is connected in series with the third resistor of 10 Ω. So, the total resistance R_total​ is:

R_total=R_eq+R₃=5 Ω+10 Ω=15 Ω

Ohm's law is given by:

V=I⋅R_total

Where:

• V=6 V
• R_total=15 Ω

Rearranging the formula to find the current III:

​$$I = \frac{V}{R_{total}} = \frac{6}{15} = 0.4 \, \text{A}$$​​

The power supplied by the battery is given by the formula:

P=V⋅I

Substituting the known values:

P=6 ×0.4 A=2.4

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