5 V battery: -1.5 V
- This gives us the equation: \(1kI_1 + 470I_2 - 1.5 = 0\)
- Loop 2: Starting from the top right corner and moving clockwise, we have:
- Voltage drop across 3.3 kΩ resistor: \(I_2 \times 3.3 \text{ kΩ}\)
- Voltage drop across 2.2 kΩ
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