Question

Three resistances are connected as shown in diagram through the resistance 5 $\Omega$, a current of 1 A is flowing :

(i) What is the current through the other two resistors?

(ii) What is the potential difference (p.d.) across AB and across AC ?

(iii) What is the total resistance ? [5 MARKS]

Answer 1

Problem: 5 Marks
First 2 points:1.5 Marks
Third point: 3 Marks

1. For current in parallel resistors
For same potential difference across two parallel resistors,

$V=I_1{R}_1=I_2{R}_2$

i.e. $\frac{I_1}{I_2}=\frac{R_2}{R_1}$

Current divides itself in the inverse ratio of the resistances.

Also total current, $I=I_1+I_2$

$\frac{I_1}{I_2}=\frac{R_2}{R_1}=\frac{15}{10}=\frac{3}{2}$

Also, $I_1+I_2=1A$

$I_1=0.6A,I_2=0.4A$

Current is 0.6 A through $10\Omega$

2. For p.d. across AB

From Ohm's law, $R=\frac{V}{I},V=IR$

$V=I\times 5=5V$

P.D. across AB

For parallel combination of $10\Omega$ and $15\Omega$ P.D. across BC, $V=I_1{R}_1=0.6\times 10=6V$

P.D. across AC = P. D across AB + P.D. across BC.

= 5 + 6 = 11 V

3. For total circuit resistance

For $10\Omega$ and $15\Omega$ in parallel

$R_p=\frac{10\times 15}{10+15}=\frac{150}{25}=6\Omega$

Total resistance $=5+6=11\Omega$

Total circuit resistance $=11\Omega$

$[Also\frac{V}{I}=\frac{11}{1}=11\Omega ]$