Question

A battery of emf 24v and internal resistance r is connected to a circuit having two parallel resistors of 3 Ohms and 6 Ohms in series with an I ohm resistor. Consider the current flowing in the 3 Ohms resistor is 0.8A, calculate the current in the 6 Ohms resistor.

Answer 1

As per the question,

Emf, E=24volt

Resistance r$_1$ =6$\Omega$

Resistance r$_2$ =3$\Omega$

current in resistance 3$\Omega$ =0.8A

Let current in 6$\Omega$ be i,

then we know In parallel circuit the voltage remains same,

so

voltage in 6$\Omega$ resistance=Voltage in 3$\Omega$ resistance

Using Ohm's law

i$\times 6=0.8\times3$

i=$\frac{0.8\times3}{6}$

i=$\frac{2.4}{6}$

i=0.4 Ampere

On solving the equation we get

i=0.4Ampere

Hence the current in 6$\Omega$ resistor is 0.4A