Question

If a 25 Ω and 15 Ω are connected in series and this combination is connected in parallel across a 40 Ω resistor, what is the effective resistance of the combination ?

Answer 1

Answer:

20 ohm's

Explanation:

Two resistance connected in series are

R=25

R'=15

total resistance =25+15= 40

now this is connected with a another 40 ohm resistor

so'

TOTAL EFFECTIVE resistance of the combination be

1/ER=1/40+1/40

1/ER=1/20

ER=20

Answer 2

Given :

• Resistors in the series circuit = 25 Ω and 15 Ω

• Resistor in the parallel circuit = 40 Ω

To find :

Equivalent resistance of the combination.

Solution :

First let us find the Total resistance in the series combination :

We know the formula for total equivalent resistance , i.e,

Rn = R1 + R2 + R3 + ... + Re

Where :

• Rn = Total resistance
• R = Resistors

Now Substituting the values in it , we get :

==> Rn = R1 + R2

==> Rn = (25 + 15) Ω

==> Rn = 40 Ω

∴ Rn = 40 Ω

Hence the Equivalent resistance in the Series circuit is 40 Ω.

Now let us find the Total resistance in the parallel circuit .

We know the formula for Total resistance in a parallel circuit i.e,

1/Rn = 1/R1 + 1/R2 + 1/R3 + ... + 1/Re

Where :

• Rn = Total resistance
• R = Resistors

Now using the formula for total resistance in parallel circuit and substituting in it, we get :

==> 1/Rn = 1/R1 + 1/R2

==> 1/Rn = (1/40 + 1/40) Ω

Here , LCM is 40.

==> 1/Rn = [(1 + 1)/40] Ω

==> 1/Rn = (2/40) Ω

==> Rn = (40/2) Ω

==> Rn = 20 Ω

∴ Rn = 20 Ω

Hence the total resistance in the parallel circuit is 20 Ω.

Since , the Total resistance in the parallel circuit is the combination of all the Resistors , the equivalent resistance in the circuit is 20 Ω.