Name: Introduction to Electric Current
Uploaded: 2019-07-19
Duration: 36 min 24 s
Description: In this video, we will learn the definition of electric current, its unit and the flow of the current in an electric circuit due to the movement of electrons.

Question

Calculate equivalent resistance between

A

and

B

of the circuit shown in figure.

Accepted Answer

Using the current distribution, we can draw the above circuit as shown below

The equivalent resistance between $E$ and $F$ :

There are two resistors of resistance $9 Ω$ are in series combination. Their equivalent resistance will be $9 + 9 = 18 Ω$ . Now $18 Ω$ is connected parallel with the $9 Ω$ . So equivalent resistance between $E and F$

$R_{EF} = \frac{9 \times 18}{9 + 18} = 6 Ω$

Equivalent resistance between $C$ and $D$ :

There are two resistors of resistance $15 Ω$ are connected in series combination and their equivalent resistance will be $R_{{CD}_{1}} = 15 + 15 = 30 Ω$

Resistance across points $C and D$ are $12 Ω, R_{EF}, 12 Ω$ are connected in a series combination. So their equivalent resistance $R_{{CD}_{2}} = 12 + 6 + 12 = 30 Ω$

Now $R_{{CD}_{1}}$ and $R_{{CD}_{2}}$ are in parallel combination. So their equivalent resistance

$R_{CD} = \frac{{R_{CD}}_{1} \times {R_{CD}}_{2}}{{R_{CD}}_{1} + {R_{CD}}_{2}} = \frac{30 \times 30}{30 + 30} = 15 Ω$

The equivalent resistance between $I$ and $J$ :

The two $3 Ω$ resistors are connected in series combination and their equivalent resistance will be $3 + 3 = 6 Ω$ . Now $6 Ω$ is connected parallel with $3 Ω$ . So, the equivalent resistance between $I$ and $J$

$R_{IJ} = \frac{6 \times 3}{6 + 3} = 2 Ω$

Equivalent resistance between $G$ and $H$ :

Two resistors of resistance $5 Ω$ are connected in series and their equivalent resistance $R_{{GH}_{1}} = 5 + 5 = 10 Ω$

Resistance across points $G$ and $H$ are $4 Ω, R_{IJ}, 4 Ω$ are connected in series combination. Their equivalent resistance $R_{{GH}_{2}} = 4 + 2 + 4 = 10 Ω$ .

Now $R_{{GH}_{1}}$ and $R_{{GH}_{2}}$ are connected in parallel combination and their equivalent resistance $R_{GH} = \frac{R_{{GH}_{1}} \times R_{{GH}_{2}}}{R_{{GH}_{1}} + R_{{GH}_{2}}} = \frac{10 \times 10}{10 + 10} = 5 Ω$

The equivalent resistance between $A$ and $B$ :

Resistors $2 Ω, R_{GH} = 5 Ω, 2 Ω$ are connected in series across the terminal $A B$ . Their equivalent resistance $R_{{AB}_{1}} = 2 + 5 + 2 = 9 Ω$

Resistors $6 Ω, R_{CD} = 15 Ω, 6 Ω$ are connected in series combination and their equivalent resistance

$R_{{AB}_{2}} = 6 + 15 + 6 = 27 Ω$ .

Now $R_{{AB}_{1}} and R_{{AB}_{2}}$ are connected in parallel combination and their equivalent resistance $R_{AB} = \frac{R_{{AB}_{1}} \times R_{AB 2}}{R_{AB 1} + R_{AB 2}} = \frac{9 \times 27}{9 + 27} = 6.75 Ω$

Calculate equivalent resistance between $A$ and $B$ of the circuit shown in figure.

Important Questions on Electric Current and Circuits

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