Question

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Answer 1

The circuit can be reduced, step by step, to a single equivalent resistance. The 8 ohm and the 8 ohm are connected in parallel , and so they can be replaced by an equivalent resistor Rp of 4 ohms, using the below equation.

Rp=

(8+8)

(8×8)

=4 ohms

This resistor is connected in series with the 4 ohm resistor R1. The total resistance Rt of the circuit is then

R

t

=R

1

+R

p

=4+4=8 ohms.

Since the 4 ohm and the 4 ohm are connected in series, they have the same current I1, which must be equal to the current of the battery. Using Ohms law we get,

I=

R

V

=

8

8

=1A.

Hence, the current flowing through the resistor R1 (4 ohms) is 1A.