question_answer

The three resistances A, B and C have values 3R, 6R and R respectively. When some potential difference is applied across the network, the thermal powers dissipated by A, B and C are in ratio

A) \[2 : 3 : 4\]                    

B) \[2 : 4 : 3\]

C) \[4 : 2 : 3\]                    

D) \[3 : 2 : 4\]

Correct Answer: C

Solution :

Thermal power in \[A={{P}_{A}}\,\,=\,\,{{\left( \frac{2i}{3} \right)}^{2}}\,3R\,\,=\,\,\frac{4}{3}\,{{i}^{2}}R\] Thermal power in \[B={{P}_{B}}\,\,=\,\,{{\left( \frac{i}{3} \right)}^{2}}\,6R\,\,=\,\,\frac{2}{3}\,{{i}^{2}}R\] Thermal power in \[C=PC={{i}^{2}}\text{R}\] \[\Rightarrow \,\,\,\,\,\,\,{{P}_{A}}:{{P}_{B}}:{{P}_{C}}=\frac{4}{3}:\frac{2}{3}:1\,\,=\,\,4:2:3\]