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DC Circuits And Measurements

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Question
ICSEENIPH12029770
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In the circuit shown in Figure 3, E1 = 17V,E2 = 21V, R1 = 2μ, R2=3μ R1=2μ R2= 3μ and R3 = 5μ.

Using Kirchoffs laws, find the currents flowing through the resistors R1,R2 and R3. (Internal resistance of each of the batteries is neglegible.)

Solution

(a) Applying Kirchoffs voltage Law to Loop ABEF

R1l1 + IR3 - E1 = 0 2I1 + 5I

                      = 17 I = I1 + l2 2I1 + 5I1 + 5l1 
                      = 17

7I1 + 5I2           = 17                          ... (i)

Applying voltage law to Loop BCDE, we have

R2I2 + R3I - 2I = 0 3I2 + 5I1 + 5I2 

                      = 21
5I1 + 8I2           = 21                    ... (ii)

Solving equations (i) and (ii)
Multiplying eq. (1) by 5 throughout

[ 7I1 + 5I2 = 17] x5  
  35I1 + 25I2 = 85                            ... (iii)
Multiplying eq. (2) by 7 throughout, we get
[ 5I1 + 8I2 = 21 ] x7  
  35I1 + 56I2 = 147                          ... (iv)

On solving equations (iii) and (iv), we get
-31 I2 = -62
This implies, 
I2 = 2 Amp, and
I1 = 1 Amp
Current in R1=I1 = 1 Amp
Current in R2 =I2 = 2 Amp
Therefore,
Current in R3 =I1+I2 = 3 Amp


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Question
ICSEENIPH12029800
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Which conservation principle is involved in Kirchoffs first law of electric circuits?

Solution

Law of Conservation of charge.

Question
ICSEENIPH12029817
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In the circuit shown in Figure 4 below, E is a battery of emf 6V and internal resistance 1Ω. Find the reading of the ammeter A, if it has negligible resistance :

Solution

Equivalent resistors across the terminal AB is given by, 
R subscript A B end subscript space equals space fraction numerator 8 space straight x space 24 over denominator 8 space plus space 24 end fraction space equals space 6 capital omega space space space i space equals space 6 over 12 space equals space 1 half space equals space 0.5 space A
 
Reading on the ammeter = 0.5 A

Question
ICSEENIPH12029818
Wired Faculty App

With the help of a neatly drawn and lebelled diagram, obtain balancing condition of a Wheatstone bridge.

Solution

Current are distributed as shown in the figure below.
Resistances P, Q, R and S are shown.
Division of current are marked.

Applying KVL to the mesh ABD,

i1P + igG - (i - i1)R = 0                                  ---(i)

Applying KVL to the mash BCD,

(ii - ig)Q - (i - ii + ig)s - igG =0                     ---(ii)

under balanced condition ig = 0

ii P = (i - ii)R                                              ---(iii)

ii Q = (i - ii)S                                             ---(iv)

Dividing eq. (iii) by eq. (iv)
straight P over straight Q space equals space R over S

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