Current Electricity

The current in the net circuit is I = $\frac{V}{R_{\text{effective}}}$ = $\frac{V}{R_1 + \frac{R_2 R_L}{R_1 + R_L}}$ = = 7.5 mA The potential across $R_L$ is $V_L$ = $I \times \frac{R_2 R_L}{R_2 + R_L}$ = 7.5 × $10^{-3}$ × 1.2 × $10^3$ = 9 V Therefore, I = $\frac{9}{R_L}$ = 6 mA The ratio of power dissipated in $R_1$ and $R_2$ is $\frac{I R_1}{(I - i) R_2}$ = $\frac{(7.5 \times 10^{-3})^2 \times 2 \times 10^3}{(1.5 \times 10^{-3})^2 \times 10^3}$ = 0.75 J The power dissipated in $R_L$ is $\frac{V_L^2}{R_L}$ = $\frac{9^2}{1.5 \times 10^3}$ = 54 mJ If $R_1$ and $R_2$ are interchanged, then I' = $\frac{V}{R_{\text{effective}}}$ = $\frac{V}{R_2 + \frac{R_1 R_L}{R_L + R_1}}$ = = 3.5 mA Hence, the voltage drop across the load is $V'_L$ = $I \times \frac{R_1 R_L}{R_1 + R_L}$ = 1 V Therefore, i′ = 2 mA. The magnitude of the power dissipated is the ratio between $i^2$ and $(i')^2$ : $\frac{P_1}{P_2}$ = $\frac{i^2}{(i')^2}$ = $\frac{6^2}{2^2}$ = 9