In the circuit shown in fig the voltage across the 2 Ω resistor is 12V. What is the emf of the battery and the current thorugh the 6 Ω resistor?

Explanation

Form Ohm's law, the voltage drop across the `6Omega` resistor is
`V =IR = (4A) (6Omega) = 24 V`
The voltage drop across the `8 Omegaa` resistor is the sum of these two
currents i.e., :7 A. The voltage drop across the `25Omega` resistro is
`V = IR = (7A)(25 Omega) = 175V`
Total voltage drop across the top branch of the circuit is
175+24 = 199V
Which is also the voltage drop across the `20Omega` resistor. THe
current through the `20Omega` resistor is then
`I = V//R = 199 V //20 Omega = 9.95A`