Question

Figure shows a network of three resistances. When some potential difference is applied across the network, thermal powers dissipated by A, B, and C are in the ratio

• A. 2 : 3 : 4
• B. 2 : 4 : 3
• C. 4 : 2 : 3
• D. 3 : 2 : 4

Answer 1

The correct option is C 4 : 2 : 3

Let $I$ be the current flow from b to a.
As resistor $3R$ and $6R$ are in parallel, so current through $3R$ is $I_3=\frac{6R}{3R+6R}I=2I/3$

and current through $6R$ is $I_6=\frac{3R}{3R+6R}I=I/3$

Thus, $P_A:P_B:P_C=I_3^2\left(3R\right):I_6^2\left(6R\right):I^{2}R=\left(2I/3{)}^2\right(3R):(I/3{)}^2\left(6R\right):I^{2}R=\frac{4}{3}:\frac{2}{3}:1=4:2:3$