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Question: In the circuit (Fig) what is potential difference \[{V_B} - {V_A}\] (in V) When current \[I\] is \[5...

In the circuit (Fig) what is potential difference VB−VA{V_B} - {V_A} (in V) When current II is 5 μA5\,\mu {\text{A}} and is decreasing at the rate of 10−3 A⋅s−1{10^{ - 3}}\,{\text{A}} \cdot {{\text{s}}^{ - 1}}?

Explanation

Solution

Use the expression for Ohm’s law. Also use the formula for emf or potential difference across the inductor. Using these two formulae, calculate the potential difference across the resistor and inductor given in the circuit. Then apply Kirchhoff’s voltage law to the circuit and then calculate the potential difference.

Formulae used:
The expression for Ohm’s law is given by
V=IRV = IR …… (1)
Here, VV is the potential difference across the ends of the conductor, II is current in the conductor and RR is the resistance of the conductor.
The emf ee induced in a inductor is given by
e=Ldidte = L\dfrac{{di}}{{dt}} …… (2)
Here, LL is the inductance of the inductor and didt\dfrac{{di}}{{dt}} is the rate of change of current with time.

Complete step by step answer:
We have given that the current in the given arrangement is 5 μA5\,\mu {\text{A}} and the rate of decreases of current is 10−3 A⋅s−1{10^{ - 3}}\,{\text{A}} \cdot {{\text{s}}^{ - 1}}.
I=5 μAI = 5\,\mu {\text{A}}
⇒didt=−10−3 A⋅s−1\Rightarrow\dfrac{{di}}{{dt}} = - {10^{ - 3}}\,{\text{A}} \cdot {{\text{s}}^{ - 1}}
We have asked to calculate the potential difference across the ends A and B of the given circuit.Let us first calculate the potential difference across the resistor.Substitute 5 μA5\,\mu {\text{A}} for II and 1 Ω1\,\Omega for RR in equation (1).
V=(5 μA)(1 Ω)V = \left( {5\,\mu {\text{A}}} \right)\left( {1\,\Omega } \right)
⇒V=(5×10−6 A)(1 Ω)\Rightarrow V = \left( {5 \times {{10}^{ - 6}}\,{\text{A}}} \right)\left( {1\,\Omega } \right)
⇒V=5×10−6 V\Rightarrow V = 5 \times {10^{ - 6}}\,{\text{V}}
Hence, the potential difference across the resistor is 5×10−6 V5 \times {10^{ - 6}}\,{\text{V}}.

Let us now calculate the emf or potential difference across the inductor,
Substitute 5 mH5\,{\text{mH}} for LL and −10−3 A⋅s−1 - {10^{ - 3}}\,{\text{A}} \cdot {{\text{s}}^{ - 1}} for didt\dfrac{{di}}{{dt}} in equation (2).
e=(5 mH)(−10−3 A⋅s−1)e = \left( {5\,{\text{mH}}} \right)\left( { - {{10}^{ - 3}}\,{\text{A}} \cdot {{\text{s}}^{ - 1}}} \right)
⇒e=(5×10−3 H)(−10−3 A⋅s−1)\Rightarrow e = \left( {5 \times {{10}^{ - 3}}\,{\text{H}}} \right)\left( { - {{10}^{ - 3}}\,{\text{A}} \cdot {{\text{s}}^{ - 1}}} \right)
⇒e=−5×10−6 V\Rightarrow e = - 5 \times {10^{ - 6}}\,{\text{V}}
Hence, the potential difference across the inductor is 5×10−6 V5 \times {10^{ - 6}}\,{\text{V}}.

Let us now determine the potential difference across the ends A and B of the given circuit.Apply Kirchhoff’s voltage law to the given circuit.
VA−V+(15 V)−e=VB{V_A} - V + \left( {15\,{\text{V}}} \right) - e = {V_B}
⇒VB−VA=−V+(15 V)−e\Rightarrow {V_B} - {V_A} = - V + \left( {15\,{\text{V}}} \right) - e
Substitute 5×10−6 V5 \times {10^{ - 6}}\,{\text{V}} for VV and 5×10−6 V5 \times {10^{ - 6}}\,{\text{V}} for ee in the above equation.
⇒VB−VA=−(5×10−6 V)+(15 V)−(−5×10−6 V)\Rightarrow {V_B} - {V_A} = - \left( {5 \times {{10}^{ - 6}}\,{\text{V}}} \right) + \left( {15\,{\text{V}}} \right) - \left( { - 5 \times {{10}^{ - 6}}\,{\text{V}}} \right)
∴VB−VA=15 V\therefore {V_B} - {V_A} = 15\,{\text{V}}

Hence, the required potential difference is 15 V15\,{\text{V}}.

Note: The students should keep in mind that we have given the rate of decreasing the current. Hence, the value of this decreasing current rate must be used with a negative sign. Also the students should be careful while using Kirchhoff’s voltage law and use the proper signs for the difference potentials.