The document describes how to calculate the voltage (V0) across a circuit with parallel branches. It can be neglected that no current flows through the 2 ohm resistor since its terminals are open. Using Kirchoff's laws, two equations are written relating the currents in each parallel branch (I1 and I2). Solving these simultaneously yields the individual currents, which are then used in Kirchoff's voltage law to calculate V0 across the output as -24.67V.

1. question 3, current divider
find v0 in the circuit below. Design a voltage divider that will produce out
Solution
Since the terminals of V o are open, no current flows through 2 ohm resistor in this branch.
Hence, it can be neglected.
There are two parallel branches here:
Branch 1, which contains two resistors - 14 ohms and 8 ohms
Branch 2, which contains three resistors - 6 ohms, 4 ohms and 10 ohms
Let current through branch 1 be I 1
Let current through branch 2 be I 2
By Kirchoff's Current Law, we know that:
I 1 + I 2 = 7
By KVL, we know that the voltage drop across the two branches will be same (since they are in
parallel).
(14 + 8) I 1 = (6 + 4 + 10) I 2
22 I 1 = 20 I 2
Solving the two equations simultaneously for I 1 and I 2 , we have:
I 1 = 10 / 3
I 2 = 11 / 3

2. We apply Kirchoff's Voltage Law by moving from positive terminal to negative terminal of V o
V o = 8 x I 1 - (10 + 4) x I 2
V o = -74 / 3 = -24.67 V