Question

Two Resistor $2\Omega$ and $4\Omega$ are connected in parallel. Two more Resistor $3\Omega$ and $6\Omega$ are connected in parallel. These two combinations are in series with a battery of emf $5V$ and internal Resistance $0.7\Omega$. Calculate the current through $6\Omega$ Resistors.

Answer 1

Total current I flowing through the circuit $I=\frac{V}{R}$

Here $R=equivalentresistance=3.03\Omega$-----refer the figure given.

$I_{total}$=$\frac{5}{3.03}$=$1.65A$

Since the resistors are in series connection (fig. B) current I will flow equally throughout the circuit.

Now calculating for voltage in resistor of $2\Omega$

$V_B$=$IR=3.30V$, then current $I$ flowing through 6$\Omega$ resistor $I_{6_{\Omega }}$=$\frac{3.30}{6}=0.55$