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#### Four resistances of $15\Omega ,12\Omega,4\Omega\:\:and\:\:10\Omega$ respectively in cyclic order to form Wheatstone's network. The resistance that is to be connected in parallel with the resistance of $10\Omega$ to balance the network is ___________$\Omega$. Option: 1 10 Option: 2 5 Option: 3 15 Option: 4 20

For balanced Wheatstone bridge $\frac{R_1}{R_2}=\frac{R_3}{R_4} \ \ or \ \ \frac{R_1}{R_3}=\frac{R_2}{R_4}$

$R_2=12 \Omega \ \ and \ \ R_4=4 \Omega$

As $\frac{R_2}{R_4}=\frac{12}{4}=3$

So using $R_1=15 \Omega$

We get $R_3=R_{AD}=5 \Omega$

let we connected x-ohms in parallel to 10-ohm resistance

i.e $R_3=5 \Omega=\frac{x*10}{x+10}$

we get $x=10 \Omega$

So the answer will be 10.

#### The length of a potentiometer wire is $1200cm$ and it carries a current of $60mA$. For a cell of emf $5V$ and internal resistance of $20\Omega$, the null point on it is found to be at $1000cm$. The resistance of whole wire is $(in \ \Omega)$:   Option: 1 100 Option: 2 80 Option: 3 604 Option: 7 120

In the potentiometer, a battery of known emf  E is connected in the secondary circuit. A constant current I is flowing through AB from the driver circuit. The jockey is a slide on potentiometer wire AB to obtain null deflection in the galvanometer. Let  $l$ be the length at which galvanometer shows null deflection.

Since the potential of wire AB (V) is proportional to the length AB(L).

Similarly $E \ \ \alpha \ \ \ l$

So we get

$\frac{V}{E}=\frac{L}{l}$

$V=E\frac{L}{l}$

$V=5\times\frac{1200}{1000}=6V$

given current through potentiometer wire =60mA

V=iR

$\\60\times10^{-3}R=6\\\Rightarrow R=100\Omega$

So the correct option is 4.

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then,

#### Two identical capacitors A and B, charged to the same potential 5V are connected in two different circuits as shown below at time t=0. If the charge on capacitors A and B at time t=CR is  and  respectively , then (Here e is the base of natural logarithm) Option: 1       Option: 2         Option: 3         Option: 4

Maximum charge on the capacitor

(a) is reverse biased and (b) is forward biased

(a)                                                           (b)

So,

Hence the correct option is (4).

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#### The current I1 (in A) flowing through $1\Omega$ resistor in the following circuit is:-                                 Option: 1 0.25  Option: 2 0.40 Option: 3 0.2   Option: 4 0.5

Ohms Law -

1. Ohm’s law

In a conductor, if all external physical conditions like temperature and pressure are kept constant the Current flowing through a conductor is directly proportional to the Potential difference across two ends.

$V\propto I$

$V=IR$

$R-$ Electric Resistance

• The graph between V and I

The slope gives the resistance

• The graph between V and I at different temperatures

Here T1>T2. The resistance of a conductor increases with increase in temperature

-

$R_{AB}=\frac{1}{2}+2=\frac{5}{2}$

$R_{CD}=2$

$R_{eq}=\frac{2.5*2}{2.5+2}=\frac{10}{9}$

$I=\frac{V}{R_{eq}}=\frac{10}{9}$

As

$\Delta V_{AB}= \Delta V_{CD}\\ \Rightarrow 2.5*I_2 =2*I_3...(1)$

and $I_2+I_3=I=\frac{10}{9 }...(2)$

From equation (1) and (2)

$I_3=\frac{50}{81} , I_2=\frac{40}{81} ,$

and $I_1= \frac{I_2}{2}=\frac{20}{81} = 0.2469$

$I_1= 0.25$  ( when considered up to two decimal point)

Or,

$I_1= 0.2$ (when considered up to one decimal point)

Answer given by the NTA is $I_1= 0.2$ But most appropriate answer is $I_1= 0.25$

#### The series combination of two batteries, both of the same emf 10 V , but different internal resistance of 20 and 5 , is connected to the parallel combination of two resistors 30 and R . The voltage difference across the battery of internal resistance 20 is zero, the value of R (in ) is __________ . Option: 1 30 Option: 2 60 Option: 3 90 Option: 4 45

Hence option (1) is correct.

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#### A galvanometer having a coil resistance 100 $\Omega$ gives a full scale deflection when a current of 1 mA is passed through it. What is the value of the resistance (in range of $k\Omega$ )which can convert this galvanometer into a voltmeter giving full-scale deflection for a potential difference of 10 V ? Option: 1 9.9 Option: 3 10 Option: 5 8.9 Option: 7 7.9

\begin{aligned} V_{g} &=i_{g} R_{g}=0.1 \mathrm{V} \\ V &=10 \mathrm{V} \\ \mathrm{R} &=\mathrm{R}_{g}\left(\frac{\mathrm{V}}{\mathrm{V}_{g}}-1\right) \\ &=100 \times 99=9.9 \mathrm{K} \Omega \end{aligned}

Hence the correct option is (1).

#### In the given circuit diagram, a wire is joining points B and D. The current in this wire is: (current in amperes,A)  Option: 1 2 Option: 2 0 Option: 3 0.4 Option: 4 4

$\\ \text{Reff} = \frac{4}{5} +\frac{6}{5} = 2\Omega \\ i = \frac{20}{2} = 10A$

So after drawing the circuit

$I = \frac{4i}{5} - \frac{3i}{5} = +\frac{i}{5} = 2A$

Hence the option correct option is (1).

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#### As shown in the figure, a battery of emf is connected to an inductor L and resistance R in series. The switch is closed at t = 0. The total charge that flows from the battery, between t = 0 and t = tc (tc is the time constant of the circuit ) is :  Option: 1 Option: 2  Option: 3  Option: 4

As current at any time t is given as

So

So integrating this will give total charge

Hence the correct option is (4).

#### An emf of 20 V is applied at time t =0 to a circuit containing in series 10mH inductor and $5\: \Omega$ resistor. The ratio of the currents at time $t=\infty$ and at $t=40\: s$ is close to: $(Take\; e^{2}=7.389)$   Option: 1 1.15 Option: 2 1.06 Option: 3 0.84 Option: 4 1.46

As we know that the -

$i = i_o(1-e^{\frac{-t}{L/R}})$

So, from the question, we can write that -

$i = \frac{20}{5}(1-e^{\frac{-t}{0.01/5}})$

$= i = 4(1-e^{{-500t}})$

By putting t = $\infty$

We get, $i_o = 4$

Again at t = 40 sec

$i_{40}= 4(1-\frac{1}{(e^2)^{10000}})$

$= 4(1-\frac{1}{(7.389)^{10000}})$

So,

$\frac{i_{\infty}}{i_{40}} \approx 1.15$

So option 2  is much closer. So option 1 will be correct.