Physics - Current Electricity Question with Solution | TestHub

Given circuit below



Resistance between$B$and$C$on the right-hand side of the circuit

$=\frac{10\times 10}{10+10}=\frac{100}{20}=5\mathrm{\Omega }$

because$1\mathrm{\Omega }+4\mathrm{\Omega }+5\mathrm{\Omega }=10\mathrm{ }\mathrm{\Omega }$are in parallel with$10 \mathrm{\Omega }$.

Resistance between$A$and$D$on the right-hand side of the circuit.

$=\frac{10\times 10}{10+10}=\frac{100}{20}=5\mathrm{\Omega }$

because$5\mathrm{\Omega }$(equivalent resistance of$1\mathrm{\Omega },\mathrm{ }4\mathrm{\Omega },\mathrm{ }5\mathrm{\Omega }$and$10 \mathrm{\Omega }$) is in series with$3\mathrm{\Omega }$and$2\mathrm{\Omega }$.

Hence, resistance between$A$and$D$will be$5\mathrm{\Omega }$

Equivalent resistance of the circuit,

$R=5+2+5=12\mathrm{\Omega }$

Current drawn from the battery,

$I=\frac{12V}{12\mathrm{\Omega }}=1\text{A}$

At the junction$\text{A}$, the current of$1 \text{A}$is divided equally between the$10\mathrm{\Omega }$resistance and the remaining circuit of the$10\mathrm{\Omega }$resistance.

At the junction$B$, the current of$0.5 \text{A}$is divided equally between the$10\mathrm{\Omega }$resistor and remaining circuit of resistance$10\mathrm{\Omega }$.

$\therefore$Current through the$4\mathrm{\Omega }$resistor.