Question

Four resistors of resistances $0.5\Omega ,1.5\Omega ,4\Omega$ and $6\Omega$ are connected in series to a battery of EMF $6V$ and negligible internal resistance. Calculate the:
(a) current drawn from the cell.
(b) potential difference at the end of each resistor.
[4 MARKS]

Answer 1

Each subpart: 2 Marks each

When the resistances are connected in series, the equivalent resistance is calculated by adding them.
Equivalent resistance of circuit,
$R=0.5+1.5+4+6=$$12\Omega$
(a) Current drawn from the cell,
$I=\frac{V}{R}=\frac{6}{12}=.5A$
(b) Potential difference across $0.5\Omega$ resistor,
$V_1=IR=0.5\times 0.5=0.25V$
Potential difference across $1.5\Omega$ resistor,
$V_2=IR=0.5\times 1.5=0.75V$
Potential difference across $4\Omega$ resistor,
$V_3=IR=0.5\times 4=2V$
Potential difference across $6\Omega$ resistor,
$V_4=IR=0.5\times 6=3V$