Question

Four resistors of resistance $0.5\Omega ,1.5\Omega ,4\Omega$and $6\Omega$ are connected in series to a battery of e.m.f. $6\mathrm{V}$ and negligible internal resistance. Calculate (a) current drawn from the cell (b) p. d. at the ends of each resistor.

Answer 1

Step 1: (a) current drawn from the cell

The equivalent resistance is derived by summing the resistance when they are connected in series.

Circuit equivalent resistance,

$R=0.5\Omega +1.5\Omega +4\Omega +6\Omega$

$R=12\Omega$

By applying ohm’s law,

$I=\frac{V}{R}$

We know that,

$V=6\text{volts},R=12\text{Ω}$

$I=\frac{6}{12}$

$I=0.5A$

Hence, the current drawn from the cell is $0.5A$.

Step 2: (b) p. d. at the ends of each resistor.

The difference in voltage between $R_1=0.5\Omega$ resistors,

$V_1=I{R}_1$

$=0.5\times 0.5$

$=0.25V$

The difference in voltage between $R_2=1.5\Omega$ resistors,

$V_2=I{R}_2$

$=0.5\times 1.5$

$=0.75V$

The difference in voltage between $R_3=4\Omega$ resistors,

$V_3=I{R}_3$

$=0.5\times 4$

$=2V$

The difference in voltage between $R_4=6\Omega$ resistors,

$V_4=I{R}_4$

$=0.5\times 6$

$=3V$

Hence, the potential difference at the end of each resistor is $V_1=0.5V,V_2=0.75V,V_3=2V,V_4=3V$.