Question

Solve it fast

in the following circuit diagram two resistors of 3 ohm and 5 ohm and a battery of 6 volt all are connected in series the potential difference across 3 ohm resistor will be

Answer 1

Given:

R1=5Ω

R2=10Ω

EMF=6V

A)

To get minimum current, the two resistors should be connected in series.

To get maximum current, we should connect the resistance in parallel

B)To calculate strength of total current in each case:

1) To get minimum current, the two resistors should be connected in series.

Effective resistance for series connection: R= R1+R2

R=5+10

R=15Ω

By Ohm's law, we have V=IR

so, I=V/R

⇒6/15⇒0.4A

I=0.4A

2) To get maximum current, we should connect the resistance in parallel.

Effective resistance for parallel connection: 1/R=1/R1+1/R2

1/R=R1+R2/R1R2

∴R=R1R2/R1+R2

⇒5x10/5+10

⇒50/15⇒10/3Ω

By Ohm's law, we have V=IR

so, I=V/R

⇒(6/10)x3⇒1.8A

I=2.25A

Hope its clear

Answer 2

In a series combination, the electric current remains same while potential difference differs at each resistor.

Here, R = R₁ + R₂

R = ( 3 + 5 ) Ω

R = 8 Ω

R = V/i

8 = 6/i

i = 6/8

i = 3/4 A

Since current remains same everwhere,

R₁ = V₁/i

3 = V₁ / (3/4)

3 × 3/4 = V₁

9/4 = V₁

2.25 V = V₁

Hence potential difference over 3 Ω resistora is 2.25 Volts (Option B).