In the circuit shown in figure, `E_1 and E_2` are two ideal sources of unknown emfs. Some currents are shown. Potential difference appearing across `6Omega` resistance is `V_(A) - V_(B) = 10V. `

Options

A. The current in the 4.00 resistor is 5 A.
B. The unknown emf `E_1` is 36 V.
C. The unknown emf `E_2 is 54 V. `
D. The resistance R is equal to

Answer

A, B, C, D

Explanation

a., b., c., d.

After redrawing the circuit

a. Current through `4Omega = 5A`
b. From loop (1)
`-8(3) + E_1 - 4(3) = 0 or E_1 = 36V`
From loop (2) : `+4(5) + 5(2) - E_2 + 8(3) = 0`
or `E_2 = 54 V`
c. From loop (3): `-2R - E_1 + E_2 = 0`
or `R= (E_2-E_1)/2 = (54-36)/2 = 9 Omega` .