Question

show how you would connected three resistors ,each of resistance 6 G, so that the combination has a resistance of (1 ) 9 resistor, (2) 4 resistor.​

Answer 1

Appropriate Question:

Show how you would connected three resistors ,each of resistance 6 Ω, so that the combination has a resistance of (1) 9 Ω, (2) 4 Ω.

Answer:

1) Connect the parallel combination of two 6 Ω resistors with one 6 Ω in a series combination.

2) Connect the series combination of two 6 Ω resistors with one 6 Ω in a parallel combination.

Explanation:

1) To get the equivalent resistance of 9 Ω :

Refer to the attachment (i).

$\rm R_1$ and $\rm R_2$ are connected in parallel combination, so their combined resistance will be :

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

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

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

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

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

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

Now, the combined resistance of $\rm R_1$ and $\rm R_2$ is connected in series combination with $\rm R_3$ . So, equivalent resistance of the circuit will be :

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

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

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

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

2) To get the equivalent resistance of 4 Ω :

Refer to the attachment (ii).

$\rm R_1$ and $\rm R_2$ are connected in series combination, so their combined resistance will be :

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

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

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

Now, the combined resistance of $\rm R_1$ and $\rm R_2$ is connected in parallel combination with $\rm R_3$ . So, equivalent resistance of the circuit will be :

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

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

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

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

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

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

∴ If we connect the series combination of two 6 Ω resistors with one 6 Ω in a parallel combination, we get the equivalent resistance of 4 Ω.