Question

3 resistances 1Ω,2Ω,4Ω are connected in series to 14V battery. What is current in 1Ω. What is p.d across 2Ω.

Answer 1

Answer:

Let, R  

1

​  

, R  

2

​  

 and R  

3

​  

 be the three resistors connected in series across a 12 V battery. I be the current in circuit.

V=IR

V=I(R  

1

​  

+R  

2

​  

+R  

3

​  

)

V=IR  

1

​  

+IR  

2

​  

+IR  

3

​  

 ..................(1)

Now, given that,

R  

1

​  

=1Ω

IR  

2

​  

=4V and

P  

3

​  

=I  

2

R  

3

​  

⇒R  

3

​  

=  

I  

2

 

P  

3

​  

 

​  

,P  

3

​  

=12W and V=12V.

Using this values in (1), we get,

12=I(1)+4+I(  

I  

2

 

12

​  

)

8=I+  

I

12

​  

 

I  

2

+12=8I

I  

2

−8I+12=0

Solving for roots we get

I−2=0 or I−6=0

I=2A or I=6A

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Explanation:

Answer 2

Explanation:

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