In the given figure,the emf of the cell is 2. | vedclass

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(C) From the circuit diagram,the four resistors are connected in parallel between points $A$ and $B$. The values of the resistors are $4 \, \Omega, 8

Vedclass · Accepted Answer

$2.2$

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(C) From the circuit diagram,the four resistors are connected in parallel between points $A$ and $B$. The values of the resistors are $4 \, \Omega, 8 \, \Omega, 12 \, \Omega$,and $6 \, \Omega$.The equivalent resistance $R_{eq}$ of the parallel combination is given by:$\frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{6}$$\frac{1}{R_{eq}} = \frac{6 + 3 + 2 + 4}{24} = \frac{15}{24} = \frac{5}{8} \, \Omega^{-1}$$R_{eq} = \frac{8}{5} = 1.6 \, \Omega$The total resistance of the circuit $R_T$ including the internal resistance $r = 0.6 \, \Omega$ is:$R_T = R_{eq} + r = 1.6 + 0.6 = 2.2 \, \Omega$The power dissipated in the whole circuit is given by $P = \frac{E^2}{R_T}$,where $E = 2.2 \, V$ is the $emf$ of the cell:$P = \frac{(2.2)^2}{2.2} = 2.2 \, W$Thus,the power dissipated in the whole circuit is $2.2 \, W$.

In the given figure,the $emf$ of the cell is $2.2 \, V$ and the internal resistance is $0.6 \, \Omega$. Calculate the power dissipated in the whole circuit (in $W$).

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