Question

............................................................??????????????

Answer 1

Answer:

$I_{1}$ = 0.7 A, $I_{2}$ = 1.3 A and ε = 7.4 V

Explanation:

From the given circuit, applying Kirchhoff's rule;

Ammeter reading, $I_{0}$ = 2 A

⇒    $I_{0}$ = $I_{1}$ + $I_{2}$ = 2 A

Dividing the circuit to loops 1 and 2.

a. From loop 1,

15 + 7$I_{1}$ - 5$I_{0}$ = 0

15 + 7$I_{1}$ - 10 = 0    (since $I_{0}$ = 2 A)

7$I_{1}$ - 5 = 0

$I_{1}$ = 0.7 A

But,  $I_{0}$ = $I_{1}$ + $I_{2}$

⇒      2 = 0.7 + $I_{2}$

        $I_{2}$ = 1.3 A

b. From loop 2,

ε + 2$I_{2}$ - 5$I_{0}$ = 0

ε + 2$I_{2}$ - 10 = 0

ε + 2.6 - 10 = 0

ε - 7.4 = 0

ε = 7.4 V

Therefore, $I_{1}$ = 0.7 A, $I_{2}$ = 1.3 A and ε = 7.4 V.