Question: Q. 1. State Kirchhoff’s laws of current distribution in an electrical network.

Using these rules determine the value of the current I1 in the electric circuit given below :

R&A [Delhi I, II, III 2014]

Show Answer

Solution:

Ans. (i) Kirchhoff’s laws 1

(ii) Finding value of current I1 in electric circuit 2

[CBSE Marking Scheme 2014]

Detailed Answer :

(i) Try yourself Similar to Q. 1, Short Answer Type Question-I

(ii) Now from the given figure,

I1+I2=I3 I2=I3I1

For ABEFA

(iii)20I140I3=40 20(I1+2I3)=40 I1+2I3=2

For BCDEB

(iv)40I3+20I2=80+40 20(2I3+I2)=120 2I3+I2=6 2I3+(I3I1)=6 2I3+I3I1=6 3I3I1=6 using eq. (i) 

Now solving eqns. (iii) and (iv),

We get

I3=85=1.6 A

and

I1=65 A=12 A

Now, from equation (ii),

I2=I3I1 =1.6(1.2) =1.6+1.2 =2.8 A

[AI Q. 2. Using Kirchhoff’s rules, calculate the potential difference between B and D in the circuit diagram as shown in the figure.

A [CBSE-2018 Comptt.]

Sol. Writing the two loop equations

1/2+1/2

Finding the current through DB

Finding the p.d. between B and D

Using Kirchhoff’s voltage rule, we have :

For loop DABD

(1×I1+(1)+(2)+2I1+2(I1+I2)=0

Or 5I1+2I2=1

..(i) 1/2

For loopDCBD

+I2×3+(3)+(1)+1×I2+2(I1+I2)=0

Or 2I1)+6I2=2

…(ii) 1/2

Solving (i) and (ii), we get

I1=513 A I2=613 A

Current through DB=I1+I2=113 A

1/2

P.D. between B and D=0.154 A

[CBSE Marking Scheme 2018]



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