Question

Calculate equivalent resistance between the points A and B.
​

Answer 1

Given :

• Resistance in the Resistor , J1 = 2 Ω
• Resistance in the Resistor, J2 = 4 Ω
• Resistor in the Resistor, J3 = 4 Ω
• Resistor in the Resistor, J4= 4 Ω
• Resistor in the Resistor, J6 = 6 Ω
• Resistor in the Resistor, J7 = 2 Ω

To find :

• Equivalent resistance in the circuit, Re = ?

Knowledge required :

• Total resistance in a parallel circuit :

$\sf{\dfrac{1}{R_{e}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dfrac{1}{R_{3}} + ... + \dfrac{1}{R_{n}}}$

• Total resistance in a series Circuit :

$\sf{R_{e} = R_{1} + R_{2} + R_{3} + ... + R_{n}}$

Where,

• Re = Equivalent resistance
• R = Resistance in the Resistors

Solution :

First let us find the resistance between the J2 , J3 and J4.

Since all the resistors are connected by a same wire , they are in series circuit.

By using the formula for resistance in a series circuit and substituting the values in it, we get :

$:\implies \sf{R_{e} = R_{1} + R_{2} + R_{3}} \\ \\ \\ :\implies \sf{R_{e} = 4 + 4 +4} \\ \\ \\ :\implies \sf{R_{e} = 12} \\ \\ \\ \boxed{\therefore \sf{R_{e} = 12\:\Omega}}$

Hence the total resistance in the J1 , J3 and J4 is 12 Ω.

Let total above Resistance be J5.

Now according to the figure ,the Resistors J5 , J6 are in the series Circuit.

So by using the formula for total resistance in the parallel circuit and Substituting the values in it, we get :

$:\implies \sf{\dfrac{1}{R_{e}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}}} \\ \\ \\ :\implies \sf{\dfrac{1}{R_{e}} = \dfrac{1}{12} + \dfrac{1}{6}} \\ \\ \\ :\implies \sf{\dfrac{1}{R_{e}} = \dfrac{1 + 2}{12}} \\ \\ \\ :\implies \sf{\dfrac{1}{R_{e}} = \dfrac{3}{12}} \\ \\ \\ :\implies \sf{R_{e} = \dfrac{12}{3}} \\ \\ \\ :\implies \sf{R_{e} = 4} \\ \\ \\ \boxed{\therefore \sf{R_{e} = 4\:\Omega}}$

Thus the total resistance between j5 and J6 is 4 Ω

Let the total above Resistance be J8.

Now we get the resistor, J1 , J8 and J7 are in series circuit , so by using the formula for total resistance in a series circuit and substituting the values in it, we get :

$:\implies \sf{R_{e} = R_{1} + R_{2} + R_{3}} \\ \\ \\ :\implies \sf{R_{e} = 2 + 4 + 2} \\ \\ \\ :\implies \sf{R_{e} = 8} \\ \\ \\ \boxed{\therefore \sf{R_{e} = 8\:\Omega}}$

Thus, the total resistance in the J1 , J8 and J7 is 8 Ω.

Hence the equivalent resistance in the circuit is 8 Ω

Answer 2

Answer:

Given :

Resistance in the Resistor , J1 = 2 Ω

Resistance in the Resistor, J2 = 4 Ω

Resistor in the Resistor, J3 = 4 Ω

Resistor in the Resistor, J4= 4 Ω

Resistor in the Resistor, J6 = 6 Ω

Resistor in the Resistor, J7 = 2 Ω

To find :

Equivalent resistance in the circuit, Re = ?

Knowledge required :

Total resistance in a parallel circuit :

\begin{gathered}\sf{\dfrac{1}{R_{e}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dfrac{1}{R_{3}} + ... + \dfrac{1}{R_{n}}} \\ \\ \end{gathered}

R

e

1

=

R

1

1

+

R

2

1

+

R

3

1

+...+

R

n

1

Total resistance in a series Circuit :

\begin{gathered}\sf{R_{e} = R_{1} + R_{2} + R_{3} + ... + R_{n}} \\ \\ \end{gathered}

R

e

=R

1

+R

2

+R

3

+...+R

n

Where,

Re = Equivalent resistance

R = Resistance in the Resistors

Solution :

First let us find the resistance between the J2 , J3 and J4.

Since all the resistors are connected by a same wire , they are in series circuit.

By using the formula for resistance in a series circuit and substituting the values in it, we get :

\begin{gathered}:\implies \sf{R_{e} = R_{1} + R_{2} + R_{3}} \\ \\ \\ :\implies \sf{R_{e} = 4 + 4 +4} \\ \\ \\ :\implies \sf{R_{e} = 12} \\ \\ \\ \boxed{\therefore \sf{R_{e} = 12\:\text{\O}mega}} \\ \\ \end{gathered}

:⟹R

e

=R

1

+R

2

+R

3

:⟹R

e

=4+4+4

:⟹R

e

=12

∴R

e

=12Ømega

Hence the total resistance in the J1 , J3 and J4 is 12 Ω.

Let total above Resistance be J5.

Now according to the figure ,the Resistors J5 , J6 are in the series Circuit.

So by using the formula for total resistance in the parallel circuit and Substituting the values in it, we get :

\begin{gathered}:\implies \sf{\dfrac{1}{R_{e}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}}} \\ \\ \\ :\implies \sf{\dfrac{1}{R_{e}} = \dfrac{1}{12} + \dfrac{1}{6}} \\ \\ \\ :\implies \sf{\dfrac{1}{R_{e}} = \dfrac{1 + 2}{12}} \\ \\ \\ :\implies \sf{\dfrac{1}{R_{e}} = \dfrac{3}{12}} \\ \\ \\ :\implies \sf{R_{e} = \dfrac{12}{3}} \\ \\ \\ :\implies \sf{R_{e} = 4} \\ \\ \\ \boxed{\therefore \sf{R_{e} = 4\:\text{\O}mega}} \\ \\ \end{gathered}

:⟹

R

e

1

=

R

1

1

+

R

2

1

:⟹

R

e

1

=

12

1

+

6

1

:⟹

R

e

1

=

12

1+2

:⟹

R

e

1

=

12

3

:⟹R

e

=

3

12

:⟹R

e

=4

∴R

e

=4Ømega

Thus the total resistance between j5 and J6 is 4 Ω

Let the total above Resistance be J8.

Now we get the resistor, J1 , J8 and J7 are in series circuit , so by using the formula for total resistance in a series circuit and substituting the values in it, we get :

\begin{gathered}:\implies \sf{R_{e} = R_{1} + R_{2} + R_{3}} \\ \\ \\ :\implies \sf{R_{e} = 2 + 4 + 2} \\ \\ \\ :\implies \sf{R_{e} = 8} \\ \\ \\ \boxed{\therefore \sf{R_{e} = 8\:\text{\O}mega}} \\ \\ \end{gathered}

:⟹R

e

=R

1

+R

2

+R

3

:⟹R

e

=2+4+2

:⟹R

e

=8

∴R

e

=8Ømega

Thus, the total resistance in the J1 , J8 and J7 is 8 Ω.

Hence the equivalent resistance in the circuit is 8 Ω