Question 17: Current Electricity

Question

The power dissipated in the circuit shown in the figure $$40 \mathrm{~W}$$. The value of $$R$$ is.

Accepted Answer

As we know,
$$\begin{aligned}
P & =\frac{V^2}{R_{\mathrm{eq}}} \
\frac{1}{R_{\mathrm{eq}}} & =\frac{1}{R}+\frac{1}{5}=\frac{5+R}{5 R} \
\Rightarrow \quad R_{\mathrm{eq}} & =\frac{5 R}{5+R} \Rightarrow P=40 \mathrm{~W}
\end{aligned}$$
$$40=\frac{10^2}{\frac{5 R}{(5+R)}} \Rightarrow R=5 \Omega$$

Answer

20$$\Omega$$

Answer

15$$\Omega$$

Answer

5$$\Omega$$

Answer

30$$\Omega$$

Current Electricity

The power dissipated in the circuit shown in the figure $40 \mathrm{~W}. The value of R$ is.

Choose the correct answer: