Question

A 5 ohms resistor is connected to two dry cells, each with Emf 3 V and internal resistance 1 ohms calculate the current through the resistor when

a) the cells are arranged in series
b) the cells are arranged in parallel​

Answer 1

Answer:

(1) when  the cells are arranged in series I = 6/7 Amp

(2)when cells are arranged in parallel​ I = 6/11 Amp

Explanation:

given a resistance 5 ohm

Two cell of emf 3 volt with 1 ohm internal resistance

(1) when  the cells are arranged in series

when two cell of emf e₁ and e₂ with internal resistance r₁ and r₂ connected in series then

equivalent emf E = e₁+e₂

E=3+3

E= 6 volt

equivalent internal resistance r = r₁+r₂

r=1+1

r= 2 ohm

R=5 ohm resistance connect with this then calculate current in circuit

current in circuit I = E/R+r

I = 6/(5+2)

I = 6/(7)

I = 6/7 Amp

(2)when cells are arranged in parallel​

when two cell of emf e₁ and e₂ with internal resistance r₁ and r₂ connected in parallel then

equivalent emf E =$\frac{e1r2+e2r1}{r1+r2}$

E =(3×1+3×1)/2

E= 6/2

E= 3 volt

net internal resistance 1/r= 1/r₁+1/₂

1/r = 1/1+1/1

1/r = 2/1

r= 1/2 ohm

now

R=5 ohm resistance connect with this then calculate

current in circuit I = E/R+r

I = 3/(5+1/2)

I = 3/(11/2)

I = 6/11 Amp

Answer 2

Answer:

R

1

and R

2

be the two resistance.

When the resistances are connected in series R

s

=R

1

+R

2

Current in the circuit is

I=

R+r

ϵ

i.e.,

5

2

=

R

s

+0

2

i.e., ⇒R

s

=R

1

+R

2

=5....(1)

When two resistances are connected in parallel

R

p

1

=

R

1

1

+

R

2

1

=

R

1

R

2

R

1

+R

2

=

R

1

R

2

5

(∵from(1))

R

p

=

5

R

1

R

2

Current in the circuit is

I=

R

p

+r

ϵ

3

5

=

(

5

R

1

R

2

)+0

2

=

R

1

R

2

10

R

1

R

2

=6....(2)

Using the relation (R

1

−R

2

)

2

=(R

1

+R

2

)

2

−4R

1

R

2

(R

1

−R

2

)

2

=(5)

2

−4(6)=25−24=1

∴R

1

−R

2

=1....(3)

Adding equations (1) and (3), we get

2R

1

=6 or R

1

=3Ω

R

2

=5−R

2

=5−3=2Ω.