Question

In the circuit shown E is a cell emf 2 V and internal resistance $0.3\Omega$. Three resistors $2\Omega ,4\Omega ,5\Omega$ are connected as shown. The power dissipated is $2\Omega$ resistor is nearly.

• A. 1.8 W
• B. 0.18 W
• C. 0.57 W
• D. 1.65 W

Answer 1

The correct option is B 0.18 W

From the given figure

$\begin{array}{c}{R}_{ab}=\frac{1}{\frac{1}{5}+\frac{1}{6}}=\frac{30}{11}\Omega \\ R_{net}=\frac{30}{11}+\frac{33}{110}=\frac{333}{110}\Omega \\ Currentproducedfrombattery=\frac{v}{R}\\ =\frac{2}{\left(\frac{333}{110}\right)}\\ =\frac{220}{333}A\\ \\ Currentinlowerbrachofab=\frac{5.1}{5+6}\\ =\frac{5}{11}.\frac{220}{333}\\ =\frac{110}{333}A\\ Powerdisspatedin2\Omega resistor=I^{2}R\\ ={\left(\frac{100}{333}\right)}^2\times 2\\ =0.18\Omega \\ Hence,optionBisthecorrectanswer.\end{array}$