Question:
The figure shows three circuits I, II and III which are connected to a 3V battery. If the powers dissipated by the configurations I, II and III are \(P_1\), \(P_2\) and \(P_3\) respectively, then :
The figure shows three circuits I, II and III which are connected to a 3V battery. If the powers dissipated by the configurations I, II and III are \(P_1\), \(P_2\) and \(P_3\) respectively, then :
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Power drawn from a fixed 3 V battery is \(V^2/R_{eq}\), so find the equivalent resistance of each circuit; the smaller the resistance, the larger the power.
Updated On: Nov 11, 2026
- \(P_1 \gt P_2 \gt P_3\)
- \(P_1 \gt P_3 \gt P_2\)
- \(P_2 \gt P_1 \gt P_3\)
- \(P_3 \gt P_2 \gt P_1\)
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The Correct Option is C
Solution and Explanation
Power from the battery: \(P = \frac{V^2}{R_{eq}}\) with \(V = 3\) V.
Circuit I: label the top node T, the node below the first 1 \(\Omega\) as J1, the node between the 2nd and 3rd vertical resistors as J2 and the bottom node as B. The resistors are T-J1, T-J2 (the right-hand 1 \(\Omega\)), J1-J2, J1-B (the left-hand 1 \(\Omega\)) and J2-B, all 1 \(\Omega\). This is a balanced bridge (J1 and J2 are at equal potential), so the J1-J2 resistor carries no current. \(R_1 = (1+1)\parallel(1+1) = 1\,\Omega\), \(P_1 = 9/1 = 9\) W.
Circuit II: the two side paths are 1+1 = 2 \(\Omega\) each and the middle resistor is 1 \(\Omega\), all in parallel: \(\frac{1}{R_2} = \frac12+\frac12+1 = 2\), \(R_2 = 0.5\,\Omega\), \(P_2 = 9/0.5 = 18\) W.
Circuit III: the diamond is (1+1)\(\parallel\)(1+1) = 1 \(\Omega\), in series with the extra 1 \(\Omega\): \(R_3 = 2\,\Omega\), \(P_3 = 9/2 = 4.5\) W.
So \(P_2 \gt P_1 \gt P_3\).
Answer: (3) \(P_2 \gt P_1 \gt P_3\)
Circuit I: label the top node T, the node below the first 1 \(\Omega\) as J1, the node between the 2nd and 3rd vertical resistors as J2 and the bottom node as B. The resistors are T-J1, T-J2 (the right-hand 1 \(\Omega\)), J1-J2, J1-B (the left-hand 1 \(\Omega\)) and J2-B, all 1 \(\Omega\). This is a balanced bridge (J1 and J2 are at equal potential), so the J1-J2 resistor carries no current. \(R_1 = (1+1)\parallel(1+1) = 1\,\Omega\), \(P_1 = 9/1 = 9\) W.
Circuit II: the two side paths are 1+1 = 2 \(\Omega\) each and the middle resistor is 1 \(\Omega\), all in parallel: \(\frac{1}{R_2} = \frac12+\frac12+1 = 2\), \(R_2 = 0.5\,\Omega\), \(P_2 = 9/0.5 = 18\) W.
Circuit III: the diamond is (1+1)\(\parallel\)(1+1) = 1 \(\Omega\), in series with the extra 1 \(\Omega\): \(R_3 = 2\,\Omega\), \(P_3 = 9/2 = 4.5\) W.
So \(P_2 \gt P_1 \gt P_3\).
Answer: (3) \(P_2 \gt P_1 \gt P_3\)
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