Source current by Kirchhoff laws

Reading the figure

Bridge: A–B = 25 Ω, B–C = 50 Ω, A–D = 20 Ω, D–C = 10 Ω, B–D = 100 Ω. The source S and a 5 Ω resistor are in series and connected between A and C (outside the bridge).

Step 1: Voltage across BD

Current 0.2 A flows from D to B, so V_D − V_B = 0.2 × 100 = 20 V.

Take V_A = 0 and let V_B = x, so V_D = x + 20 and V_C = c.

Step 2: KCL at B and D

At B (in from A, in from D, out to C): (0 − x)/25 + 0.2 = (x − c)/50 ⇒ c = 3x − 10 ... (1)

At D (in from A; out to B and to C): (0 − (x + 20))/20 = 0.2 + (x + 20 − c)/10 ⇒ c = 1.5 x + 32 ... (2)

From (1) and (2): 3x − 10 = 1.5 x + 32, so x = 28 V and c = 74 V.

So V_B = 28 V, V_D = 48 V, V_C = 74 V (with V_A = 0).

Step 3: Bridge currents

• B to A: 28/25 = 1.12 A
• D to A: 48/20 = 2.4 A
• C to B: (74 − 28)/50 = 0.92 A
• C to D: (74 − 48)/10 = 2.6 A

Current entering the bridge at C = 0.92 + 2.6 = 3.52 A, and the same leaves at A.

Step 4: Source current and polarity

The source branch must carry 3.52 A from A back to C, through the 5 Ω and S.

Source current = 3.52 A

Voltage needed: C is at +74 V relative to A. The 5 Ω drops 5 × 3.52 = 17.6 V, so the source must supply E = 74 + 17.6 = 91.6 V

Polarity: the positive terminal of S is on the C side (towards C), the negative terminal is towards the 5 Ω and A. The source current is 3.52 A and the EMF is about 91.6 V.