Current Electricity Test

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Current Electricity Test - 81

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Weekly Quiz Competition

Question 1
1 / -0

The electromotive force resistance of a single cell equivalent to a combination of $$6$$ cells as shown in the figure will be:-

A
$$4 V, 1.5 \Omega$$

B
$$6 V, 1.5 \Omega$$

C
$$8 V, 1.5 \Omega$$

D
$$0 V, 1.5 \Omega$$

Solution
Question 2
1 / -0

In a experiment for calibration of voltmeter a standard cell of e.m.f $$1.5 \ V$$ is balanced at $$300 \ cm$$ length of potentiometer wire. The potential difference across a resistance in the circuit is balanced at $$1.25 \ m$$. If a voltmeter is connected across the same resistance it reads $$0.65 \ V$$. The error in the volt meter is:

A
$$0.5 \ V$$

B
$$0.025 \ V$$

C
$$0.05 \ V$$

D
$$0.25 \ V$$

Solution
Question 3
1 / -0

Two cells of electro motive force $$3 \ V$$ and $$5 \ V$$ and internal resistance $$r_1$$ and $$r_2$$ respectively are in series with an external resistance $$R$$. If the potential difference across 1st cell is zero, then $$R$$ is:-

A
$$\dfrac{5 r_1 - 3 r_2}{3}$$

B
$$\dfrac{2 r_1 - 3 r_2}{4}$$

C
$$\dfrac{3 r_1 - 5 r_2}{3}$$

D
$$\dfrac{4 r_1 - 5 r_2}{3}$$

Solution
Question 4
1 / -0

A coil of area $$500{cm}^{2}$$ having $$1000$$ turns is held perpendicular to a uniform magnetic field $$0.4G$$. The induced emf in the coil if it is turned through angle of $${180}^{o}$$ in $$\cfrac{1}{10}s$$ is

A
$$0.01V$$

B
$$0.02V$$

C
$$0.03V$$

D
$$0.04V$$

Solution
Question 5
1 / -0

An AC supply is connected to a resistor. When the peak value of the $$e.m.f.$$ of the supply is $$V_0$$ and the frequency is $$f$$, the mean power disspated in the resistor is $$P$$. The supply frequency is then changed to $$2f$$, the peak value of the $$e.m.f.$$ remaining as $$V_0$$ what is now the mean power in the resistor?

A
P

B
$$\sqrt2$$P

C
$$2P$$

D
$$4P$$
Question 6
1 / -0

The emf of a celll is $$E$$, and its internal resistance is $$1\ \Omega$$. A resistance of $$4\Omega$$ is joined to battery in parallel this is connected in secondary circuit of potentio meter the balancing length is $$160\ cm$$. If $$1V$$ cell balances for $$100\ cm$$ of potentio meter wire, the emf of cell $$E$$ is

A
$$1V$$

B
$$3V$$

C
$$2V$$

D
$$4V$$

Solution
Question 7
1 / -0

A conducting loop is placed in a uniform magnetic field of induction $$B$$ tesla with its plane normal to the field. Now, the radius of the loop starts shirking at the rate $$(\dfrac{dr}{dt})$$.
Then the induced emf at the instant when the radius is $$r$$ is

A
$$ \pi rB \left(\dfrac{dr}{dt} \right)$$

B
$$ 2 \pi rB \left(\dfrac{dr}{dt} \right)$$

C
$$ \pi r^2 \left(\dfrac{dB}{dt} \right)$$

D
$$\left(\dfrac{\pi r^2}{2}\right)^2 B\left(\dfrac{dr}{dt}\right)$$

Solution
Question 8
1 / -0

A thin uniform ring of radius R carrying uniformly distributed charge Q and mass M rotates about its axis with angular velocity . The ratio of its magnetic moment and angular momentum is .

A
$$\frac{Q}{M}$$

B
$$\frac{M}{Q}$$

C
$$\frac{1}{2}\frac{Q}{M}$$

D
$$\frac{1}{2}\frac{M}{Q}$$

Solution
Question 9
1 / -0

In he circuit shown, the current in the $$1\Omega $$ resistor is :

A
1.3 A, from P to Q

B
0A

C
0.13 A, from Q to P

D
0.14 A, from P to Q

Solution
Question 10
1 / -0

The current i in the circuit (see figure) is :

A
$$\dfrac {1}{45} A$$

B
$$\dfrac {1}{15} A$$

C
$$\dfrac {1}{10} A$$

D
$$\dfrac {1}{5} A$$

Solution

R.E.F.Image.
Apply KVL Law from ODAB
$$ 2-30i_{1}=0$$
$$ i_{1}= 2/30 = 1/15 AMP$$
Now apply KVL in ODA CBO
$$ 2-30(i-i_{1})-30(i-i_{1})=0$$
$$ 2 = 60 (i-i_{1})$$
$$ \frac{1}{30}=(i-i_{1})$$
$$ i = \frac{1}{15} = \frac{1}{30}$$
$$ i = \frac{1}{30}+\frac{1}{15}= \frac{3}{30}$$
$$ i = \frac{1}{10}AMP$$