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# A circuit element is placed in a closed box. At a time, t=0, a constant current generator supplying current of 1 amp, is connected across the box. Potential difference across the box varies according to the graph shown in figure. The element in the box has:(a) resistance of 2 ohms(b) battery of emf 6V (c) inductance of 2H (d) capacitance of 0.5F

Last updated date: 09th Aug 2024
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Hint:
1.Potential difference is rising or changing with time. And current is constant, therefore, circuit elements can’t be resistance and inductance.
2. Potential difference is changing so the battery emf can’t be constant 6 voltage. It should be time varying.

Formula Used:
1.For a purely resistive circuit element. we know, V=IR …… (1)
Where,
V = voltage across resistor
R=resistance in the circuit
I = current in the circuit
2. For a purely inductive circuit we know, \$ - I\dfrac{{di}}{{dt}} = Voltage\$ …… (2)
Where,
I = inductance across the inductor
V = Voltage across inductive coil
3. For a purely capacitive circuit we know, \$\dfrac{{dq}}{{dt}} = C\dfrac{{dV}}{{dt}}\$ \$ \Rightarrow \dfrac{i}{C} = \dfrac{{dV}}{{dt}}\$ …… (3)
Where,
Q = charge stored in capacitor
V = Voltage across plates
i = current in the circuit
C= capacitance

Let’s find the correct by rejecting the wrong choices: Hit and trial approach:

Step 1:
From equation (1) we can say with time, Voltage would be constant. Since there is no time term in it. Therefore, the above graph should have 0 slope. But slope is non zero. So, it can’t be a resistor.
\[ \Rightarrow \]Hence, option (a) Not Possible

Step 2:
Given Voltage is time varying with constant slope. But in option (b) it says emf is constant= 6V
\[ \Rightarrow \]Hence, option (b) Not Possible

Step 3:
From equation (2), we can say with time, current is also changing. But in question, it is given current is constant and equals 1 amp. Therefore, circuits can’t be inductive.
\[ \Rightarrow \]option (b) Not Possible

Step 4:
From equation (3), we can say with time Voltage is varying and current is constant. Therefore, conditions satisfied for a capacitive circuit. So, Circuit is capacitive.

Step 5:
From the graph, we can see the graph is constant. So, slope of graph=\$\dfrac{i}{C} = \dfrac{{dV}}{{dt}}\$
Evaluating slope of graph using 2-point slope form we get,
\$\dfrac{i}{C} = \dfrac{{{y_2} - {y_1}}}{{{t_2} - {t_1}}} = \dfrac{{8 - 2}}{{3 - 0}} = 2\$ where \$({x_1},{y_1})\$ and \$({x_2},{y_2})\$ are two separate points ……. (4)

Step 6:
Substitute value of current i=1 amp, in equation (4) we get,
\$\dfrac{1}{2} = C \Rightarrow C = 0.5F\$ required capacitance of circuit.