Electrostatic Potential and Capacitance Test

Result

Electrostatic Potential and Capacitance Test - 91

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Weekly Quiz Competition

Question 1
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Five capacitor each of capacity C are joined as shown in the following figure. If their resultant capacity $$ C_R = 2 \mu F $$ then the capacity of each capacitor is

A
$$ 5 \mu F $$

B
$$ 20 \mu F $$

C
$$ 10 \mu F $$

D
$$ 4 \mu F $$

Solution
Question 2
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What is the equivalent capacitance of the combination shown in the figure below.

A
$$C/2$$

B
$$C$$

C
$$2\, C$$

D
$$4\, C$$

Solution
Question 3
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Charge of 2Q and -Q are placed on two plates of a parallel plate capacitor if capacitance of capacitor is C find potential difference between the plates:

A
$$ V = \dfrac {Q}{C} $$

B
$$ V = \dfrac {3Q}{2C} $$

C
$$ V = \dfrac {2Q}{3C} $$

D
$$ V = \dfrac {2Q}{C} $$

Solution
Question 4
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In the circuit shown below $$C_1 = 10\, \mu F, C_2 = C_3 = 20\, \mu F$$, and $$C_4 = 40\, \mu F$$. If the charge on $$C_1$$ is $$20\, \mu C$$ then potential difference between $$X$$ and $$Y$$ is

A
$$2\, V$$

B
$$3\, V$$

C
$$6\, V$$

D
$$3.5\, V$$

Solution
Question 5
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The parallel combination of two air filled parallel plate capacitors of capacitance C and NC is connected to a battery of voltage, V. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is?

A
$$\dfrac{V}{K+n}$$

B
$$V$$

C
$$\dfrac{(n+1)V}{(K+n)}$$

D
$$\dfrac{nV}{K+n}$$

Solution

After fully charging, battery is disconnected.
Total charge of the system$$=CV+nCV$$
$$=(n+1)CV$$
After the insertion of dielectric of constant K
New potential (common)
$$V_C=\dfrac{total charge}{total capacitance}$$
$$=\dfrac{(n+1)CV}{KC+nC}=\dfrac{(n+1)V}{K+n}$$.
Question 6
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Two capacitors of capacitance of $$6 uF$$ and $$12 uF$$ are connected in series with a battery. The voltage across the $$6uF$$ capacitor is $$2 V$$. Compute the total battery voltage.

A
$$1V$$

B
$$2V$$

C
$$3V$$

D
$$4V$$

Solution

$$\textbf{Step 1 : Make circuit diagram}$$ $$\textbf{[Ref. Fig.1]}$$

$$\textbf{Step 2 : In series combination charge on both capacitor will be same.}$$ $$\textbf{[Ref. Fig.2]}$$
So, $$Q_1 = Q_2 = Q=C_1V_1$$
$$=6\mu F\times 2V=12\,\mu C$$

$$\textbf{Step 3 : Apply KVL in loop}$$ $$\textbf{[Ref. Fig. 2]}$$
While applying KVL in clockwise direction, increase in voltage is considered as positive and decrease in voltage as negative.
$$V-\dfrac{Q}{C_1}-\dfrac{Q}{C_2}=0$$

$$V=\dfrac{Q}{C_1}+\dfrac{Q}{C_2}=\dfrac{12\,\mu C}{6\,\mu F}+\dfrac{12\,\mu C}{12\,\mu F}$$

$$=(2+1)V=\,3V$$

Hence total battery voltage is $$3V$$

Option $$C$$ is correct.
Question 7
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Four capacitors are connected as given in figure find $$ V_a - V_b $$

A
-5 V

B
-9 V

C
-10 V

D
-13 V

Solution
Question 8
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The equivalent capacitance between $$A$$ and $$B$$ is

A
$$\dfrac {4C}{3}$$

B
$$\dfrac {8C}{3}$$

C
$$12\ C$$

D
$$\dfrac {5C}{12}$$
Question 9
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Directions For Questions

Capacitor $$C_2$$ in the circuit is a variable capacitor (its capacitance can be varied). Graph is plotted between potential difference $$V_1$$ (across capacitor $$C_1$$) versus $$C_3$$. Electric potential $$V_1$$ approaches on asymptote of $$10\ V$$ as $$C_3 \to \infty$$.

...view full instructions

The capacitance of the capacitor $$C_1$$ has value:-

A
$$2\mu F$$

B
$$6\mu F$$

C
$$8\mu F$$

D
$$12\mu F$$
Question 10
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The equivalent capacitance between points A and B is

A
$$\dfrac{3C}{4}$$

B
$$\dfrac{C}{3}$$

C
$$\dfrac{3C}{2}$$

D
None of these.

Solution