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Question: In the figure, \(\epsilon = 12.0 V,\; R_1 = 2000\Omega,\; R_2 = 3000\Omega,\; R_3 = 4000\Omega\). Wh...

In the figure, ϵ=12.0V,  R1=2000Ω,  R2=3000Ω,  R3=4000Ω\epsilon = 12.0 V,\; R_1 = 2000\Omega,\; R_2 = 3000\Omega,\; R_3 = 4000\Omega. What are the potential differences (a) VAVBV_A – V_B (b) ) VBVCV_B – V_C (c) ) VCVDV_C – V_D and (d) ) VAVCV_A – V_C ?

Explanation

Solution

In order to find the potential differences across the given terminals, we need to find the current flowing through the resistances across the terminals. For this, we need only two tools. At either node B or C apply Kirchhoff’s Current Law (KCL) to obtain current flowing through R3R_3 in terms of current flowing through R1R_1 and R2R_2.
Then, apply Kirchhoff’s Voltage Law (KVL) to the two loops EACDE and EABCDE and obtain simultaneous expressions for the currents through R1R_1 and R2R_2 and solve to determine the numerical values of currents through the different resistances. To this end, use ohm’s law to determine the voltage drops across the resistances across the given terminals and arrive at the appropriate potential differences for the same.

Formula used: Kirchhoff’s Current Law (KCL): Ientering+Iexiting=0I_{entering} + I_{exiting} = 0
Kirchhoff’s Voltage Law (KVL): ΣV=0\Sigma V =0
Ohm’s law: V=IRV=IR

Complete step by step answer:
Let us begin by deconstructing the circuit and analysing the current flow through it.
The current flows from the positive terminal to the negative terminal of the battery via the resistance setup across nodes A and D. At node A, the current splits up into two to flow through resistances R1R_1 and R2R_2. Let the current flowing through these resistances be i1i_1 and i2i_2 respectively.
Since the resistances seem to be following a symmetry in the way they are arranged, we can say that the current flowing through them can also be extrapolated under the same symmetry.
Since we have RAB=RCD=R1R_{AB} = R_{CD} =R_1, we can say that the same current i1i_1 flows through both the resistances. Similarly, since RAC=RBD=R2R_{AC} = R_{BD} =R_2, we can say that the same current i2i_2 flows through both the resistances.

Now, at node B, the current again splits up into two branches to flow through R3R_3 and R2R_2. Let the current flowing through R3R_3 be i3i_3.
At the node B, we can apply Kirchhoff’s Current Law (KCL) which suggests that the algebraic sum of currents entering a node will be equal to the sum of currents leaving a node. From the diagram, we see that at node B, applying KCL gives us:
i1=i3+i2i3=i1i2i_1 = i_3+i_2 \Rightarrow i_3 = i_1-i_2. Thus, the current flowing through R3R_3 is the difference between the currents flowing through R1R_1 and R2R_2.
Now, let us bring in Kirchhoff’s Voltage Law (KVL) which suggests that the sum of all voltages around any closed loop in a circuit must be equal to zero. Since our circuit consists of two independent closed loops, let us apply KVL and see how it goes from there.
Applying KVL in the external loop EACDE of our circuit we get:
ϵi2R2i1R1=0\epsilon -i_2R_2-i_1R_1 =0
Given that ϵ=12  V\epsilon=12\;V, R1=2  kΩR_1 = 2\;k\Omega, R2=3  kΩR_2 = 3\;k\Omega
12i2(3  kΩ)i1(2  kΩ)=0\Rightarrow 12 – i_2(3\;k\Omega)-i_1( 2\;k\Omega)=0
2000i1+3000i2=12\Rightarrow 2000i_1+3000i_2 =12
Similarly, applying KVL to the loop EABCDE we get:
ϵi1R1i3R3i1R1=0\epsilon – i_1R_1 -i_3R_3-i_1R_1=0
ϵ2i1R1(i1i2)R3=0\Rightarrow \epsilon-2i_1R_1 – (i_1-i_2)R_3 = 0
Given that R3=4  kΩR_3 = 4\;k\Omega
122i1(2  kΩ)(i1i2)(4  kΩ)=0\Rightarrow 12 – 2i_1(2\;k\Omega)-(i_1-i_2)(4\;k\Omega)=0
8000i1+4000i2=122000i1+1000i2=3\Rightarrow -8000i_1 + 4000i_2=-12 \Rightarrow -2000i_1+1000i_2=-3
Solving the two KVL equations simultaneously we get:
4000i2=9i2=940004000i_2 = 9 \Rightarrow i_2=\dfrac{9}{4000}
i2=2.25  mA\Rightarrow i_2= 2.25\;mA
Substituting this is the first KVL equation we get:
2000i1+3000(2.25×103)=122000i1=126.75i1=5.2520002000i_1+3000(2.25 \times 10^{-3}) =12 \Rightarrow 2000i_1 = 12-6.75 \Rightarrow i_1 = \dfrac{5.25}{2000}
i1=2.625  mA\Rightarrow i_1 = 2.625\;mA
Then, i3=i1i2=2.6252.25=0.375  mAi_3 = i_1-i_2 = 2.625-2.25=0.375\;mA
Now that we have found the current through each resistance, we can now find the required potential differences which are nothing but the voltage drops across the resistances in the corresponding branches.
(a) VAVB=i1R1=(2.625×103)×(2×103)=5.25  VV_A – V_B = i_1R_1 = (2.625 \times 10^{-3}) \times (2\times 10^3) = 5.25\;V
(b) VBVC=i3R3=(0.375×103)×(4×103)=1.5  VV_B – V_C = i_3R_3 = (0.375 \times 10^{-3}) \times (4\times 10^3) = 1.5\;V
(c) VCVD=i1R1=(2.625×103)×(2×103)=5.25  VV_C – V_D = i_1R_1 = (2.625 \times 10^{-3}) \times (2\times 10^3) = 5.25\;V
(d) VAVC=i2R2=(2.25×103)×(3×103)=6.75  VV_A – V_C = i_2R_2 = (2.25 \times 10^{-3}) \times (3\times 10^3) = 6.75\;V

Note: Though we looked at KCL and KVL from a quantitative perspective, it is important to understand what they mean in a physical sense.
KCL signifies conservation of charge since the law basically suggests that the sum of currents entering a node must be equal to the sum of currents leaving the node, which means that electric charges are neither ambiguously lost nor mysteriously added but remains the same in an isolated system.
KVL signifies conservation of energy since the total energy in a system remains constant, though it may be transferred between components of the system in the form of electric potential and current.