Current Electricity Test

Result

Current Electricity Test - 66

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The equivalent resistance between the points A and B is:

A
$$\dfrac{36}{7} \Omega$$

B
$$10 \Omega$$

C
$$\dfrac{85}{7} \Omega$$

D
none of these

Solution
Question 2
1 / -0

What will be the effective resistance between the points P and Q in the following circuit

A
$$ 13 \Omega$$

B
$$2/3 \Omega$$

C
$$ 2 \Omega$$

D
$$7 \Omega$$

Solution

The three resistors between P and Q are connected in series with each other.
Therefore, equivalent resistance $$=2+3+2=7 \Omega$$
Question 3
1 / -0

The $$V-I$$ graph for a conductor at temperature $$T_{1}$$ and $$T_{2}$$ are as shown in figure. The term $$(T_{2}-T_{1})$$ is proportional to:

A
$$\cos 2\theta$$

B
$$\sin 2\theta$$

C
$$\cot 2\theta$$

D
$$\tan 2\theta$$

Solution

$$\textbf{Step1: Slope of $$V-I$$ graph}$$
Ohm's Law states that $$V=IR$$
From the graph
Slope of $$V-I $$ graph will be resistance
So, $$ \dfrac{{{V}_{1}}}{{{I}_{1}}}\ =\tan \theta \ = \ {{R}_{1}} $$
$$\&$$ $$\dfrac{{{V}_{2}}}{{{I}_{2}}}\ =\tan (90-\theta )\ =\cot \theta \ =\ {{R}_{2}} $$

$$\textbf{Step2: Calculation of}$$ $$T_2 -T_1$$
Temperature dependence of resistance is given by:
$$R_1 = R_0(1+\alpha T_1)$$
$$R_2 = R_0(1+\alpha T_2)$$
Subtracting above equations, we get
$$R_2-R_1 = \alpha(T_2-T_1)$$
$$\therefore T_2-T_1$$ is proportional to $$R_2-R_1 $$
$$\Rightarrow$$ $${{T}_{2}}-{{T}_{1}}\ \propto \ \left( \cot \theta -\tan \theta \right)=\dfrac{\cos \theta }{\sin \theta }-\dfrac{\sin \theta }{\cos \theta }=\dfrac{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{\sin \theta \cos \theta } $$
$$ {{T}_{2}}-{{T}_{1}}\ \propto \ \dfrac{2\ \cos 2\theta }{\sin 2\theta }\ =\ 2\cot 2\theta $$
Hence, $${{T}_{2}}-{{T}_{1}}\ \propto \ \cot 2\theta $$
Therefore, Option C is correct
Question 4
1 / -0

$$9$$ Identical wires each of Resistance $$R$$ are connected to form a closed polygon. The equivalent resistance between the ends of any side is

A
$$9R$$

B
$$R/9$$

C
$$8R/9$$

D
$$9R/8$$
Question 5
1 / -0

$$AB$$ is part of a circuit as shown, that absorbs energy at a rate of $$50 \ W$$. $$E$$ is an electromotive force device that has no internal resistance.

A
Potential difference across $$AB$$ is $$48 V$$

B
Electromotive force device is $$48 V$$.

C
Point $$B$$ is connected to the positive terminal of $$E$$.

D
Rat of conversion from electrical to chemical energy is $$48 W$$ in device $$E$$.

Solution

$$\begin{array}{l} { i^{ 2 } }\left( 2 \right) +i.E=50 \\ \Rightarrow 2+E=50 \\ \Rightarrow E=48V \\ Hence, \\ option\, \, B\, is\, \, correct\, \, answer. \end{array}$$
Question 6
1 / -0

Potentiometer is based on

A
Deflection method

B
Zero deflection method

C
Both (a) and (b)

D
None of these

Solution
Question 7
1 / -0

Two cells each of electromotive force $$E$$ and internal resistance $$r$$ are concerned in parallel across the resistance $$R$$. The maximum energy given to the resistor per second if:

A
$$R = \dfrac{r}{2}$$

B
$$R = r$$

C
$$R = 2r$$

D
$$R = 0$$

Solution

$${R_{eq}} = \dfrac{r}{2} + R = \dfrac{{r + 2R}}{2}$$

$$I = \dfrac{{2E}}{{r + 2R}}$$

For max. power consumption$$.$$ I should be max. so denominator should be min. for that
$$r + 2R = {\left( {\sqrt r - \sqrt {2R} } \right)^2} + 2\sqrt r \sqrt {2R} $$
$$ \Rightarrow \sqrt r - \sqrt {2R} = 0$$
$$ \Rightarrow R = \dfrac{r}{2}$$
Hence,
option $$(A)$$ is correct answer.
Question 8
1 / -0

In wheatstone's bridge $$P = 9$$ ohm, $$ R = 4 $$ ohm and $$ S = 6 $$ ohm. How much resistance must be put in parallel to the resistance S to balance the bridge?

A
$$24$$ ohm

B
$$13.5$$ ohm

C
$$7.2$$ ohm

D
$$18.7$$ ohm

Solution

ACC, to what stone's bridge
At balance condition
$$PS=QR$$
$$\dfrac{PS}{R}=Q$$
$$\dfrac{9\times 6}{4}=Q=\dfrac{54}{4}$$
$$\boxed {Q=13.5\Omega}$$
Question 9
1 / -0

Two identical batteries, each of electromotive force $$2 V$$ and internal resistance $$r = 1 \Omega$$ are connected as shown. The maximum power that can be developed across $$R$$ using these batteries is:

A
$$2 W$$

B
$$3.2 W$$

C
$$8,2 W$$

D
$$4 W$$

Solution
Question 10
1 / -0

$$n$$ identical cells each of electromotive force $$e$$ and internal resistance $$r$$ are connected in series of this combination. The current through $$V$$ is:

A
$$\frac{{2e}}{n}$$

B
$$\dfrac{n \space e}{nR + r}$$

C
$$\dfrac{e}{R + nr}$$

D
$$\dfrac{n \space e}{R + r}$$

Solution

$$I = \left( {\dfrac{{ne - 2e}}{{nr}}} \right)$$
$$I = \dfrac{{\left( {n - 2} \right)e}}{{nr}}$$
$$V = e - Ir$$
$$V = e - \dfrac{{\left( {n - 2} \right)e}}{n}$$
$$V = \dfrac{{\left( {ne - ne + 2e} \right)}}{n}$$
$$V = \dfrac{{2e}}{n}$$
Hence,
option $$(A)$$is correct answer.