Current Electricity Test

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Current Electricity Test - 55

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Question 1
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A wire has resistance of $$\displaystyle 12\Omega $$. It is cut into two parts and both halve are connected in parallel. The new resistance is

A
$$\displaystyle 3\Omega $$

B
$$\displaystyle 1.5\Omega $$

C
$$\displaystyle 12\Omega $$

D
$$\displaystyle 6\Omega $$

Solution

As the wire is bent at its mid point, thus resistance of each segment $$R = \dfrac{12}{2} = 6\Omega$$
Both are in parallel
$$\therefore$$ Effective resistance between A and B $$R' = \dfrac{R_1\times R_2}{R_1+R_2} = \dfrac{6\times 6}{6+6} = 3\Omega$$
Question 2
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Two metallic wires of $$\displaystyle 6\Omega $$ and $$\displaystyle 3\Omega $$ are in connection. What will be the mode of connection so that to get effective resistance of $$\displaystyle 2\Omega $$ ?

A
Parallel

B
Series

C
Both

D
None

Solution

$$R_1$$=$$6\Omega$$ , $$R_2$$=$$3\Omega$$
Case (1): In Series connection :
$$R_{eq}$$=$$R_1$$ + $$R_2$$
$$R_{eq}$$=$$6$$ + $$3$$
$$R_{eq}$$=$$9\Omega$$
So series connection not possible.

Case (2): In Parallel connection :
$$\frac{1}{R_{eq}}$$=$$\frac{1}{R_1}+\frac{1}{R_2}$$
$$\frac{1}{R_{eq}}$$=$$\frac{1}{6}+\frac{1}{3}$$
$$R_{eq}$$ = $$\frac{6\times3}{6+3}$$= $$\frac{18}{9}$$=$$2\Omega$$
Hence the two resistors are connected in parallel.
Question 3
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Two resistors are connected a) in series b) in parallel.
The equivalent resistance in two cases are $$9\Omega$$ and $$2\Omega$$ respectively. Then the resistance of the component resistors are.

A
$$2\Omega$$ and $$7\Omega$$

B
$$3\Omega$$ and $$6\Omega$$

C
$$3\Omega$$ and $$9\Omega$$

D
$$5\Omega$$ and $$4\Omega$$

Solution

$$\textbf{Step 1: Equivalent resistance for series connection}$$
Let the resistances be $$R_1$$ and $$R_2$$.
$$R_{s} = R_1+R_2$$
$$\Rightarrow$$ $$9 = R_1+R_2$$
$$\Rightarrow R_1 = 9-R_2$$    $$......(1)$$

$$\textbf{Step 2: Equivalent resistance for parallel connection} $$
$$R_{p} = \dfrac{R_1R_2}{R_1+R_2}$$
$$\Rightarrow$$ $$2 = \dfrac{R_1 R_2}{R_1 +R_2}$$   $$......(2)$$

$$\textbf{Step 3: Solving Equations}$$
Put the value of $$R_1$$ from $$(1)$$ in equation $$(2)$$
$$2 = \dfrac{(9-R_2) R_2}{9}$$

$$\Rightarrow R_2^2 - 9R_2 +18 = 0 $$
$$\Rightarrow (R_2-6)(R_2 -3)= 0$$
$$\Rightarrow R_2= 6\Omega $$ or $$ R_2 =3\Omega$$

From equation (1)
$$\Rightarrow R_1= 3\Omega $$ or $$ R_1 =6\Omega$$

Hence, option(B) is correct.
Question 4
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A wire of resistance $$12\ ohm$$ per meter is bent to form a complete circle of radius $$10\ cm$$. The resistance b/w its two diametrically opposite points $$A$$ and $$B$$ as shown in the figure is

A
$$\displaystyle 6\pi \Omega $$

B
$$\displaystyle 3\Omega $$

C
$$\displaystyle 0.6\pi \Omega $$

D
$$\displaystyle 6\Omega $$

Solution

Length of semi-circle $$L = \pi r = \pi (0.1)$$
$$\therefore$$ Resistance of each semi-circle $$R = 12\times 0.1\pi = 1.2\pi \Omega$$
Equivalent resistance between point A and B $$R_{eq} = 1.2\pi\parallel 1.2\pi = \dfrac{1.2\pi}{2} = 0.6\pi \Omega$$
Question 5
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For three resistors connected in series.

A
$$\displaystyle R_{eff}={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }$$

B
$$\displaystyle V={ V }_{ 1 }={V }_{ 2 }={ V}_{ 3 }$$

C
$$\displaystyle \dfrac {1}{R_{eff}}=\dfrac {1}{{ R }_{ 1 }}+\dfrac {1}{ {R }_{ 2 }}+\dfrac {1}{ {R }_{ 3 }}$$

D
None of these

Solution

Same current flows through the resistors connected in series but potential difference is different across individual resistors and sum of individual potential difference is equal to the total potential difference.
$$\therefore$$ $$V = V_1+V_2+V_3$$
OR $$IR_{eff} = IR_1+IR_2+IR_3$$
$$\implies$$ $$R_{eff} = R_1+R_2+R_3$$
Question 6
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An electricity bulb of 100 watt is connected to supply of electricity of 220V. Resistance of filament is

A
$$\displaystyle 484\Omega $$

B
$$\displaystyle 100\Omega $$

C
$$\displaystyle 22000\Omega $$

D
$$\displaystyle 242\Omega $$

Solution

Given : $$V =220$$ volts $$P =100$$ W
$$\therefore$$ Resistance of the filament $$R = \dfrac{V^2}{R}$$
$$\implies$$ $$R = \dfrac{220^2}{100} = 484\Omega$$
Question 7
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When two resistances $$\displaystyle { R }_{ 1 }$$ and $$\displaystyle { R }_{ 2 }$$ are connected in series, they consume 12W power. When they are connected in parallel, they consume 50W power. What is the ratio of the powers of $$\displaystyle { R }_{ 1 }$$ and $$\displaystyle { R }_{ 2 }$$

A
$$1/4$$

B
$$4$$

C
$$3/2$$

D
$$3$$

Solution

Let the powers of the resistances be $$P_1$$ and $$P_2$$.
Parallel connection : $$P_{eq} = P_1+P_2$$
$$\therefore$$ $$50 = P_1+P_2$$ $$\implies P_2 = 50-P_1$$
Series connection : $$P_{eq} = \dfrac{P_1P_2}{P_1+P_2}$$
$$\therefore$$ $$12 = \dfrac{P_1(50 - P_1)}{50}$$
OR $$P_1^2 - 50P_1 + 600 = 0$$ $$\implies$$ $$P_1 = 30$$ W
$$\implies$$ $$P_2 = 50 - 30 = 20$$ W
Thus ratio of powers $$\dfrac{P_1}{P_2} = \dfrac{3}{2}$$
Question 8
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In the circuit, $$\displaystyle 5\Omega $$ resistor develops 45J/s due to current flowing through it. The power developed per second across $$\displaystyle 12\Omega $$ resistor is:

A
16W

B
192W

C
36W

D
64W

Solution

Voltage across $$5\Omega$$ $$V = \sqrt{PR} = \sqrt{45\times 5} = 15$$ volts
$$\therefore$$ Current $$I_1 = \dfrac{15}{5} = 3$$ A
Equivalent resistance of series combination of $$3\Omega$$ and $$6\Omega$$ $$R' = 9+6 = 15\Omega$$
$$\therefore$$ Current $$I_2 = \dfrac{15}{15} = 1$$ A
Thus total current flowing through $$12\Omega$$ $$I = I_1+I_2 = 3+1 = 4$$ A
$$\therefore$$ Power developed across $$12\Omega$$ $$P = I^2 (12) = 4^2\times 12 = 192$$ W
Question 9
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If $$\displaystyle { R }_{ 1 }$$ and $$\displaystyle { R }_{ 2 }$$ are respectively the filament resistance of a $$200 \ W$$ bulb and $$100 \ W$$ bulb designed to operate on the same voltage then

A
$$\displaystyle { R }_{ 1 }$$ is two times $$\displaystyle { R }_{ 2 }$$

B
$$\displaystyle { R }_{ 2 }$$ is two times $$\displaystyle { R }_{ 1 }$$

C
$$\displaystyle { R }_{ 2 }$$ is four times $$\displaystyle { R }_{ 1 }$$

D
$$\displaystyle { R }_{ 1 }$$ is four times $$\displaystyle { R }_{ 2 }$$

Solution

Given :
$$P_1 = 200$$ W
$$P_2 = 100$$ W

Power, voltage and resistance are related as
$$P=\dfrac{V^2}{R} \Rightarrow R=\dfrac{V^2}{P}$$

For constant voltage,
$$\implies$$ $$R\propto \dfrac{1}{P}$$

$$\therefore$$ $$\dfrac{R_1}{R_2} = \dfrac{P_2}{P_1} = \dfrac{100}{200}$$ $$\implies R_2 = 2R_1$$
Question 10
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Three resistors each having the same resistance are connected in parallel. Their equivalent resistance is $$\displaystyle 5\ \Omega $$. If they are connected in series, their equivalent resistance is.

A
$$\displaystyle 3\ \Omega $$

B
$$\displaystyle 45\ \Omega $$

C
$$\displaystyle 5\ \Omega $$

D
$$\displaystyle 8\ \Omega $$

Solution