# Question #4c100

Apr 12, 2017

$0.496 \Omega$

#### Explanation:

Let $R$ be resistance of motor.

A motor has coils turning in a magnetic field.
We know that whenever a coil turns in a magnetic field an emf is induced in the coil. This emf, known as the back emf, acts against the applied voltage that's initially caused the motor to spin.

Hence, net Voltage applied to the motor $= \text{Total emf of batteries"-"back emf}$
$= 6.0 - 4.50 = 1.50 V$

Total resistance in the circuit$= \text{Total internal resistance of batteries"+"Resistance of motor}$
$= 4 \times 0.001 + R$
$= \left(0.004 + R\right) \Omega$
Applying Ohm's law $I = \frac{V}{R}$ to the motor circuit we get
$3.0 = \frac{1.50}{0.004 + R}$
$\implies \left(0.004 + R\right) = \frac{1.50}{3.0}$
$\implies \left(0.004 + R\right) = 0.5$
$\implies R = 0.5 - 0.004$
$\implies R = 0.496 \Omega$