Correct Option: D
From the given information we can make the equivalent circuit as shown below:
Applying KVL
10 – 4.7kΩ × I – 0.7 – 2.2 kΩ × I – 5 = 0
| or I = | | 10 – 0.7 – 5 | =
| 4.3 V | = 0.623 mA
|
(4.7 + 2.2) kΩ | 6.9 Ω |
V2 = I × 2.2 kΩ = 0.623 mA × 2.2 kΩ
or V2 = 1.371 V
