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Battery emf from voltage drop on 8Ω

Short answer

With 20 V across 8 Ω, total current is 2.5 A. Parallel 11 Ω and 28 Ω branch gives 7.897 Ω. Ammeter current = 2.5 × 11/(11+28) = 0.705 A. Total resistance = 26.897 Ω, so battery EMF = 2.5 × 26.897 = 67.2 V.

A battery of unknown emf is connected across resistances, as shown in figure below. The voltage drops across the 8 Ω resistor is 20 V. What will be the current reading in the ammeter? What is the emf of the battery? [Figure: battery E with a voltmeter across the 8 Ω resistor; 11 Ω resistor in series; 15 Ω and 11 Ω resistors in a delta-like arrangement; 12 Ω resistor in series with an ammeter A in the return path.]

Reading the figure

The battery E feeds, in series: the 8 Ω resistor (with the voltmeter across it), an 11 Ω resistor, and then a parallel group. The parallel group has an 11 Ω branch and a branch made of 15 Ω, the ammeter A and 13 Ω in series. The ammeter is ideal (zero resistance).

Step 1: Total (battery) current

I = V₈/8 = 20/8 = 2.5 A

Step 2: Parallel group

Branch 1: 11 Ω. Branch 2: 15 + 13 = 28 Ω (ammeter branch).

R_p = 11 × 28/(11 + 28) = 308/39 = 7.897 Ω

Voltage across the group = 2.5 × 7.897 = 19.74 V

Step 3: Ammeter reading

I_A = 19.74/28 = 0.705 A (Check by current division: 2.5 × 11/39 = 0.705 A.)

Step 4: EMF

Total resistance = 8 + 11 + 7.897 = 26.897 Ω

E = I × R_total = 2.5 × 26.897 = 67.24 V

Answer: Ammeter reading ≈ 0.705 A; EMF of the battery ≈ 67.2 V

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