For the circuit shown in the figure,the power | vedclass

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Question

(B) The circuit consists of a $6 \, V$ battery in series with a $1.5 \, \Omega$ resistor and a parallel combination of three resistors ($4 \, \Omega$,

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) The circuit consists of a $6 \, V$ battery in series with a $1.5 \, \Omega$ resistor and a parallel combination of three resistors ($4 \, \Omega$,$4 \, \Omega$,and $6 \, \Omega$).First,calculate the equivalent resistance $(R_p)$ of the three parallel resistors:$\frac{1}{R_p} = \frac{1}{4} + \frac{1}{4} + \frac{1}{6} = \frac{3+3+2}{12} = \frac{8}{12} = \frac{2}{3} \, \Omega^{-1}$So,$R_p = 1.5 \, \Omega$.The total effective resistance of the circuit is $R_{eff} = 1.5 \, \Omega + R_p = 1.5 \, \Omega + 1.5 \, \Omega = 3 \, \Omega$.The power delivered by the battery is given by $P = \frac{V^2}{R_{eff}}$.Substituting the values,$P = \frac{(6)^2}{3} = \frac{36}{3} = 12 \, W$.

For the circuit shown in the figure,the power delivered by the battery is ................ $W$.

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