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Question: The given figure shows a part of an electrical circuit. The potentials at the points \(a,b\) and \(c...

The given figure shows a part of an electrical circuit. The potentials at the points a,ba,b and cc are 30V,12V30V,12V and 2V2V respectively. Find the currents through the three resistors.

Explanation

Solution

We have to consider the node as dd and then suppose a potential of VV exists there. As bb and cc are in parallel so the current from aa splits up into two equal values. We have to assume a current is flowing in the circuit aa. Then by comparing the equations with the help of Kirchhoff’s Law we will find the value of the unknown potential. After that we will be using Ohm’s Law to find the values of current.

Complete step by step solution:

Let the potential at point dd as shown in figure be Vd{V_d}. Let a current ii flows from aa to dd.
Since, dd to bb circuit and the dd to cc circuit are in parallel so both have the same current flowing through it.
It means that current ii is being split up into two equal magnitudes of current i2\dfrac{i}{2}. So, i2\dfrac{i}{2} current is flowing through the parallel circuit.
The potentials at the points a,ba,b and cc are 30V,12V30V,12V and 2V2V respectively.
So, Va=30V{V_a} = 30V, Vb=12V{V_b} = 12V and Vc=2V{V_c} = 2V.
And the given resistances are, Ra=10Ω{R_a} = 10\Omega , Rb=20Ω{R_b} = 20\Omega and Rc=30Ω{R_c} = 30\Omega
Now applying Kirchhoff’s Law we get,
VaVd(i×Ra)=0{V_a} - {V_d} - \left( {i \times {R_a}} \right) = 0
30Vd=10i(1)\Rightarrow 30 - {V_d} = 10i - - - - - \left( 1 \right)
Similarly,
Vd12=20×i2=10i(2){V_d} - 12 = 20 \times \dfrac{i}{2} = 10i - - - - - \left( 2 \right)
Vd2=30×i2=15i(3){V_d} - 2 = 30 \times \dfrac{i}{2} = 15i - - - - - \left( 3 \right)
Comparing equation (1)\left( 1 \right) and (2)\left( 2 \right) we get,
30Vd=Vd1230 - {V_d} = {V_d} - 12
Vd=25\therefore {V_d} = 25
So, the potential at point dd is 25V25V.
Now, we know that the currents flowing through resistances Ra{R_a}, Rb{R_b} and Rc{R_c} is ii, i2\dfrac{i}{2} and i2\dfrac{i}{2} respectively.
Now by using Ohm’s Law we get,
VaVd=i×R{V_a} - {V_d} = i \times {R_{}}
3025=i×1030 - 25 = i \times 10
So the value of i=0.5 Ai = 0.5{\text{ }}A
Therefore, i2=0.25 A\dfrac{i}{2} = 0.25{\text{ }}A.
So, the currents flowing through resistances Ra{R_a}, Rb{R_b} and Rc{R_c} is 0.5 A0.5{\text{ }}A, 0.25 A0.25{\text{ }}A and 0.25 A0.25{\text{ }}A.

Note: It must be noted that we must consider a common potential point in this type of question. As the potential at point dd is greater than bb and cc so, the current flows outwards from dd with respect to bb and cc but as the potential at point aa is greater than dd, the currents flows inwards or towards dd.