Four lamps are connected as shown in the figu | vedclass

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(D) When $S_2$ is open and $S_1$ is at position $2$,the circuit is a series combination of $a, b, c, d$. Since $P = I^2 R$,and $b$ is brightest,$R_b$

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$a, d, c, b$

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(D) When $S_2$ is open and $S_1$ is at position $2$,the circuit is a series combination of $a, b, c, d$. Since $P = I^2 R$,and $b$ is brightest,$R_b$ is the largest. Since $c$ and $d$ are dimmest and equal,$R_c = R_d$. Given the series circuit,$P_b > P_a > P_c = P_d$,which implies $R_b > R_a > R_c = R_d$.When $S_2$ is closed and $S_1$ is at position $1$,the circuit configuration changes. Lamps $b$ and $c$ are now in parallel,and this combination is in series with $a$ and $d$. Let the voltage across the parallel combination be $V_0$. Then $P'_b = V_0^2 / R_b$ and $P'_c = V_0^2 / R_c$. Since $R_b > R_c$,we have $P'_c > P'_b$.For lamps $a$ and $d$,they carry the total current $I$. Since $R_a > R_d$,$P'_a = I^2 R_a > P'_d = I^2 R_d$. Comparing the parallel branch with the series components,the power in the parallel branch is $P'_c = I_2^2 R_c = [I (R_b / (R_b + R_c))]^2 R_c$. Since $R_b > R_c$,the effective resistance of the parallel part is less than $R_d$,leading to $P'_a > P'_d > P'_c > P'_b$.

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