The current from the battery in the circuit d | vedclass

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(A) First,simplify the right side of the circuit. The resistors $7\,\Omega$,$1\,\Omega$,and $10\,\Omega$ are in series,so their equivalent resistance

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(A) First,simplify the right side of the circuit. The resistors $7\,\Omega$,$1\,\Omega$,and $10\,\Omega$ are in series,so their equivalent resistance is $R_s = 7 + 1 + 10 = 18\,\Omega$.This $18\,\Omega$ resistor is in parallel with the $6\,\Omega$ resistor. Their equivalent resistance $R_p$ is given by $\frac{1}{R_p} = \frac{1}{6} + \frac{1}{18} = \frac{3+1}{18} = \frac{4}{18}$,so $R_p = \frac{18}{4} = 4.5\,\Omega$.Now,the total resistance of the circuit is $R_{total} = 2\,\Omega + 0.5\,\Omega + 4.5\,\Omega + 8\,\Omega = 15\,\Omega$.The current from the battery is $i = \frac{V}{R_{total}} = \frac{15\,V}{15\,\Omega} = 1\,A$.

The current from the battery in the circuit diagram shown is .............. $A$.

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