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 the ratio
Medium
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(c)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 $C = {P_C} = {i^2}R$
$ \Rightarrow {P_A}:{P_B}:{P_C}$
$ = \frac{4}{3}:\frac{2}{3}:1 = 4:2:3$
Thermal power in $B = {P_B} = {\left( {\frac{i}{3}} \right)^2}6R = \frac{2}{3}{i^2}R$
Thermal power in $C = {P_C} = {i^2}R$
$ \Rightarrow {P_A}:{P_B}:{P_C}$
$ = \frac{4}{3}:\frac{2}{3}:1 = 4:2:3$
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