In the diagram above, the ratio of he electric power dissipated in the 6Ω and the 3Ω resistor respectively is
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Correct Answer: Option D
Explanation:
Power, P = I2R;
and I = E/R
But effective Parallel resistance: is given by:1/R = 1/6 + 1/3 = 1+2/6 = 3/6 => R = 6/3 = 2Ω
Current I = E/R = 12/2 = 6A
But I1 = (R1/R[sub]1+R2[/sub]) x 1 = 3/9 x 6 = 2A
again I1 = (R1/R[sub]1+R2[/sub]) x 1 = 6/9 x 6 = 4A
∴P1 = I12 R1 = 2 x 2 x 6 = 24W
P2 = I22 R2 = 4 x 4 x 3 = 48W
thus P1/P[sub]2[/sub] = 24/48 = 1/2 ; OR P1 : P2 = 1:2
Power, P = I2R;
and I = E/R
But effective Parallel resistance: is given by:1/R = 1/6 + 1/3 = 1+2/6 = 3/6 => R = 6/3 = 2Ω
Current I = E/R = 12/2 = 6A
But I1 = (R1/R[sub]1+R2[/sub]) x 1 = 3/9 x 6 = 2A
again I1 = (R1/R[sub]1+R2[/sub]) x 1 = 6/9 x 6 = 4A
∴P1 = I12 R1 = 2 x 2 x 6 = 24W
P2 = I22 R2 = 4 x 4 x 3 = 48W
thus P1/P[sub]2[/sub] = 24/48 = 1/2 ; OR P1 : P2 = 1:2