Question

In the circuit shown in Fig. E26.17, the voltage across the 2.00 Ω resistor is 12.0 V. What are the emf of the battery and the current through the 6.00 Ω resistor?

Answer 1

The emf of the battery of the given circuit is 18 volts and the current through the 6 ohm resistor is 3A.

What is ohm's law?

In a current carrying conductor, the voltage difference V is directly proportional to the current I through the conductor. The constant is called the resistance R.

i.e V=IR.

Emf of the battery is equal to e = V1 +V2

Current through 2 ohm resistor, I2 = V2 /R

I2 = 12/2

I2 =6 A

Voltage through 1 ohm resistor V1 = I1 R

V1 = 6 x 1 =6 volts

Emf of the battery E = 6+12 =18 volts.

The current  through 6 ohm resistor is

I = 18 /6 = 3 A

Thus, emf of the battery is 18 volts and the current through the 6 ohm resistor is 3A.

Learn more about ohm's law.

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