Question

Two resistor of resistance 3ohm and 6ohm respectively are connected to a battery of 6v so as to have
a. Minimum resistance
b. Minimum current
c. How will you connect the resistance in each case
d. Calculate the strength of the current in the circuit in both cases

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=1.8A

Hope its clear..