Question

show how would you connect three resistors each of resistance 6 ohm , so a combination has a resistance of a [9] b[4]

Answer 1

Explanation:

As per the provided information in the given question, we have to connect three resistors each of resistance 6 Ω in such a way that equivalent resistance becomes 9 Ω in first case and 4 Ω in second case.

(Refer to attachment for the diagram.)

• Diagram 1 : To get the equivalent resistance of 9 Ω.
• Diagram 2 : To get the equivalent resistance of 4 Ω.

To get the equivalent resistance of 9 Ω :

If we connect the one 6 Ω resistor in series combination with the two 6 Ω resistors in parallel combination, we'll get the the equivalent resistance of 9 Ω.

Quick Check!

$\bf R_1$ & $\bf R_2$ are connected in parallel combination, so combined resistance of $\bf R_1$ & $\bf R_2$ will be given by,

$\longmapsto\rm{\dfrac{1}{R_{(1,2)}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} }$

$\longmapsto\rm{\dfrac{1}{R_{(1,2)}} =\Bigg ( \dfrac{1}{6} + \dfrac{1}{6} \Bigg ) \; \Omega }$

$\longmapsto\rm{\dfrac{1}{R_{(1,2)}} =\Bigg ( \dfrac{1 + 1}{6} \Bigg ) \; \Omega }$

$\longmapsto\rm{\dfrac{1}{R_{(1,2)}} =\Bigg ( \dfrac{2}{6} \Bigg ) \; \Omega }$

On reciprocating both sides,

$\longmapsto\rm{R_{(1,2)} =\Bigg ( \dfrac{6}{2} \Bigg ) \; \Omega }$

$\longmapsto\rm{R_{(1,2)} = 3 \; \Omega }$

Now, the combined resistance of $\bf R_1$ & $\bf R_2$ is connected with $\bf R_3$ in series combination. So, equivalent resistance will be given by,

$\longmapsto\rm{R_{(1,2,3)} = R_{(1,2)} + R_3 }$

$\longmapsto\rm{R_{(1,2,3)} = (3+6)\; \Omega }$

$\longmapsto\bf {R_{(1,2,3)} = 9 \; \Omega }$

Hence, verified!

$\rule{200}2$

To get the equivalent resistance of 4 Ω :

If we connect the one 6 Ω resistor in parallel combination with the two 6 Ω resistors in series combination, we'll get the the equivalent resistance of 9 Ω.

Quick Check!

$\bf R_1$ & $\bf R_2$ are connected in series combination, so combined resistance of $\bf R_1$ & $\bf R_2$ will be given by,

$\longmapsto\rm { R_{(1,2)} = R_1 + R_2}$

$\longmapsto\rm {R_{(1,2)} = (6+6) \; \Omega }$

$\longmapsto\rm {R_{(1,2)} = 12 \; \Omega }$

Now, the combined resistance of $\bf R_1$ & $\bf R_2$ is connected with $\bf R_3$ in parallel combination. So, equivalent resistance will be given by,

$\longmapsto\rm{\dfrac{1}{R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} }$

$\longmapsto\rm{\dfrac{1}{R_{(1,2,3)}} =\Bigg ( \dfrac{1}{12} + \dfrac{1}{6} \Bigg ) \; \Omega }$

$\longmapsto\rm{\dfrac{1}{R_{(1,2,3)}} =\Bigg ( \dfrac{1 + 2}{12} \Bigg ) \; \Omega }$

$\longmapsto\rm{\dfrac{1}{R_{(1,2,3)}} =\Bigg ( \dfrac{3}{12} \Bigg ) \; \Omega }$

On reciprocating both sides,

$\longmapsto\rm{R_{(1,2,3)} =\Bigg ( \dfrac{12}{3} \Bigg ) \; \Omega }$

$\longmapsto\bf{R_{(1,2,3)} = 4 \; \Omega }$

Hence, verified!