In the given circuit,calculate the value of c | vedclass

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(D) First,calculate the net electromotive force $(E_{net})$ and net internal resistance $(r_{net})$ of the battery combination. The batteries are conn

Vedclass · Accepted Answer

$0.5$

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(D) First,calculate the net electromotive force $(E_{net})$ and net internal resistance $(r_{net})$ of the battery combination. The batteries are connected in series with opposing polarities. Thus,$E_{net} = 8\,V - 4\,V = 4\,V$. The internal resistances are in series,so $r_{net} = 0.5\,\Omega + 1.0\,\Omega = 1.5\,\Omega$.Next,calculate the equivalent resistance of the external circuit. The $3\,\Omega$ and $6\,\Omega$ resistors are in parallel,so their equivalent resistance $R_p = \frac{3 	imes 6}{3 + 6} = \frac{18}{9} = 2\,\Omega$. This is in series with the $4.5\,\Omega$ resistor,so the total external resistance $R_{ext} = 4.5\,\Omega + 2\,\Omega = 6.5\,\Omega$.The total resistance of the circuit is $R_{total} = R_{ext} + r_{net} = 6.5\,\Omega + 1.5\,\Omega = 8.0\,\Omega$.Finally,the total current in the circuit is $I = \frac{E_{net}}{R_{total}} = \frac{4\,V}{8.0\,\Omega} = 0.5\,A$. Since the $4.5\,\Omega$ resistor is in the main branch,the current flowing through it is $0.5\,A$.

In the given circuit,calculate the value of current in the $4.5\,\Omega$ resistor in $A$.

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