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In the circuit shown below the op-amp is ideal. If βF = 60, then the total current supplied by the 15 V source is:
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- 123.1 mA
- 98.3 mA
- 49.4 mA
- 168 mA
Correct Option: C
Given
V+ = 5 V
V– = VE = 5 V (due to virtual ground)
| Ie = | = 50 mA | |
| 100 |
since ideal op-amp draws no current, so
| Iz = | , Ie = Ib + Ic | |
| 47 kΩ |
| = | , or Ic = Ie – Ib | |
| 47 kΩ |
= 0·213 mA , or βIb + Ib = Ie
| Ie = | , or Ib = | ⇒ Ic = | ||||
| 100 | 1 + β | (1 + β) |
= 50 mA
The total current supplied by the 15 V source
| = Iz + Ic = Iz + | ||
| (1 + β) |
| = |  | 0·213 + |  | mA = 49.39 mA | |
| 61 |
Hence alternative (C) is the correct choice.
