Question

11. Show how you would connect three resistors, each of resistance 6 92, so that the
How much current would flow through the 12 a resistor?
combination has a resistance of (1) 9
, (ii) 42​

Answer 1

Answer:

To obtain a combined resistance of 9 ohms, three resistors each of resistance 6 ohm should be connected in a parallel circuit and one in series.

In a parallel circuit, the equivalent resistance of all the resistors connected is equal to the reciprocal value of the sum of the individual resistances.

In a series circuit, the equivalent resistance of all the resistors connected is equal to the sum of the resistance values of each of the individual resistor.

So in a parallel circuit, total resistance = \frac{1}{Rt}=\frac{1}{R 1}+\frac{1}{R 2}+\frac{1}{R 3}

Rt

1

=

R1

1

+

R2

1

+

R3

1

or

R t= \frac{\frac{1}{2}}{R 1}+\frac{1}{R 2}+\frac{1}{R 3}

R1

2

1

+

R2

1

+

R3

1

.

In a series circuit, total resistance = R1 +R2 + R3

(i) To get 9 ohms, two resistors should be connected in parallel and one in series.

So,

Rt in parallel will be

Rt = \frac{1}{\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}}=\frac{1}{\frac{1}{6}+\frac{1}{6}}

R

1

1

+

R

2

1

1

=

6

1

+

6

1

1

= 3 ohms

One resistor is connected in series with the 3 ohms, so

Rt in series will be Rt = R1 +R2 = 6+3 = 9 ohms

(ii) To obtain a combined resistance of 4 ohms, two resistors (of 6 ohms each) are connected in series and one in parallel to the series.

So,

Rt in series = R1 +R2 = 6+6 = 12 ohms

Rt in parallel = \frac{1}{R t}=\frac{1}{R 1}+\frac{1}{R 2}=\frac{1}{6}+\frac{1}{12}

Rt

1

=

R1

1

+

R2

1

=

6

1

+

12

1

= 4 Ohms