Question

In the figure shown each resistor is of $20\Omega$ and the cell has emf $10volt$ with negligible internal resistance. Then rate of joule heating in the circuit is (in watts)

• A. $100/11$
• B. $10000/11$
• C. $11$
• D. $Noneofthese$

Answer 1

The correct option is C $11$

Here three resistances are in parallel on right branch. The equivalent resistance

is $R_{eq1}=(1/20+1/20+1/20{)}^{-1}=(3/20{)}^{-1}=20/3\Omega$

On top side, two resistance are in parallel and the equivalent resistance is

$R_{eq2}=(1/20+1/20{)}^{-1}=10\Omega$

Now $R_{eq1}$ and $R_{eq2}$ are in series.

So, $R_{eq3}=R_{eq1}+R_{eq2}=20/3+10=50/3\Omega$

Net effective resistance of the circuit is $R_{eq}=(1/20+1/R_{eq3}{)}^{-1}=(1/20+3/50{)}^{-1}=100/11\Omega$

Thus, rate of Joule heating $=\frac{V^2}{R_{eq}}=\frac{{10}^2}{100/11}=11W$