Question

Question 3 For Figure 1 use Kirchhoff's Laws to find the currents 11, 12 and 13, and find the total resistance of the circuit. Total 10 marks Space to answer Question 3 here RI R; 492 1 1, 192 + + 12 22 B 28 V R2 7V B2 Figure 1:

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Answer 1

Q3)

R Х R3 412 13 112 12 + + B = 28 VL1 28 VL1 22 R, L1 L1 7V B2 T Figure 1:

There are 2 Kirchoff's Laws:

Kirchhoffs First Law – The Current Law, (KCL)

The algebraic sum of ALL the currents entering and leaving a node must be equal to zero.

For example at point X current entering is I2 and leaving are I1 and I3.

hence, I2 = I1 + I3  let this be equation 1.

Kirchhoff's Voltage Law (KVL) the algebraic sum of all the voltages around any closed loop in a circuit is equal to zero.

We take voltage rise as positive and voltage drop as negative. This means that while going according to the loops if we are moving from negative terminal to positive terminal of a voltage source we take it as positive also if we are moving opposite to the direction of current then also it is taken as positive. In cases opposite to these Voltage is taken as negative.

let us consider the given example,

In loop L1 we are going from negative to positive terminal of battery and opposite to the direction of current so they will be taken as positive and net voltage will put equal to zero.

According to Loop L1

28 + 4.I1 + 2.I2 = 0 let this be equation 2.

In loop L2 we are moving in direction of the current and from positive to negative terminal of the battery so they will be taken as negative.

-7 + (-2.I2) + (-1.I3) = 0 let this be equation 3.

Now we have to solve 3 equations for I1, I2 and I3.

Let I1, I2 and I3 be represented by x,y and z respectively just for simplicity of representation.

3 equations are:

y = x + z

-7 - 2y - z =0

=> 7 + 2y + z = 0

28+4x+2y=0.

Substitute to z from first equation into second equation:

7 +2y+y-x=0

=> 7 +3y-x =0

Multiplied this with 4

28 +12y-4x =0

Add above equation and the third equation

56+14y=0

y= -4 Ampere

Use this in 2nd equation

7 - 2(4)+z=0

z=1 Ampere

Use this in 1st equation.

-4 = x + 1

x = -5 Ampere.

Currents the negative sign implies that there are actually in opposite direction.

looking at the current will find which resistors are in parallel and which one are in series to find the net resistance.

This shows that 1 Ω and 2 Ω resistance are in parallel and their net resistance is in series with 4Ω resistance.

R, 5A X 1A R; 412 132 + + 4АУ B = 28 VL1 28 VL1 212 R, L1 7V B2 T Figure 1:

Total Resistance = 4.67 Ω

1 x2 1+2 +4 4.66666666667

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