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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.

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Posted by

vishal kumar

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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Posted by

vishal kumar

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In a meter bridge experimental S is a standard resistance, R is resistance wire. It is found that balancing length is l=25 cm. If R is replaced by a wire of half length and half diameter that of R of same material, then the balancing distance {l}' (in cm) will now be _____.
Option: 1 40
Option: 2  50
Option: 3 20
Option: 4 30
 

 

 

 

\frac{X}{R}=\frac{75}{25}=3

R=\frac{P^{l}}{A}=\frac{4P^{l}}{\pi d^{2}}

{R}'=\frac{4\rho \left ( \frac{l}{2} \right )}{\pi \left ( \frac{d}{2} \right )^{2}}=2R

then, \frac{x}{{R}'}=\frac{X}{2R}=\frac{3}{2}

l=40.00\; cm

 

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Posted by

avinash.dongre

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 Q_{A} and Q_{B} respectively , then (Here e is the base of natural logarithm)
Option: 1 Q_{A}=VC,\; Q_{B}=CV      
Option: 2  Q_{A}=\frac{CV}{2},\; Q_{B}=\frac{VC}{e}      
Option: 3   Q_{A}=\frac{VC}{e},\; Q_{B}=\frac{CV}{2}     
Option: 4  Q_{A}=VC,\; Q_{B}=\frac{VC}{e}      
 

 

 

Maximum charge on the capacitor=5CV

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

(a)                                                           (b)

                 

So, q=q_{max}\left [ e^{-t/RC} \right ]

Q_{B}=\frac{CV}{e}

Hence the correct option is (4). 

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Posted by

avinash.dongre

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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

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Posted by

Ritika Jonwal

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

 

 

 

 

 

\begin{array}{l}{\mathrm{V}_{1}=\varepsilon_{1}-\mathrm{i} . \mathrm{r}_{1}} \\ {0=10-\mathrm{i} \times 20} \\ {\mathrm{i}=0.5 \mathrm{A}} \\ {\mathrm{V}_{2}=\varepsilon_{2}-\mathrm{ir}_{2}} \\ {=10-0.5 \times 5} \\ {\mathrm{V}_{2}=7.5 \mathrm{V}} \\ {\mathrm{O} \cdot 5=\frac{7.5}{30}+\frac{7.5}{\mathrm{x}}} \\ {\frac{7.5}{x}=0.25} \\ {x=\frac{7.5}{0.25}=30}\end{array}

Hence option (1) is correct. 

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Posted by

vishal kumar

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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).

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Posted by

vishal kumar

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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Posted by

avinash.dongre

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As shown in the figure, a battery of emf \epsilon 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 \frac{\epsilon L }{R^{2}} \left ( 1 - \frac{1}{e} \right )
Option: 2 \frac{\epsilon L }{R^{2}}


Option: 3 \frac{\epsilon R }{eL^{2}}

Option: 4 \frac{\epsilon L }{eR^{2}}
 

 

 

 

 

As current at any time t is given as

I = I_{0}\left ( 1 - e^{\frac{-t}{\tau }} \right ) = \frac{\epsilon }{R}\left ( 1 - e^{\frac{-t}{\tau }} \right )

So \mathrm{q}=\int_{0}^{\mathrm{T}_{\mathrm{C}}} \mathrm{idt}

 

So integrating this will give total charge

 

q=\frac{\varepsilon}{\mathrm{R}}\left[\mathrm{t}-\frac{\mathrm{e}^{-t / \tau }}{\frac{-1}{ \tau }}\right]_{0}^{ \tau } ; \quad=\frac{\varepsilon}{\mathrm{R}}\left[ \tau + \tau \mathrm{e}^{-1}- \tau \right]

 

q=\frac{\varepsilon}{\mathrm{R}} \times \frac{1}{\mathrm{e}} \times \frac{\mathrm{L}}{\mathrm{R}} \quad ; \quad q=\frac{\varepsilon \mathrm{L}}{\mathrm{R}^{2} \mathrm{e}}

Hence the correct option is (4).

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Posted by

vishal kumar

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.

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Posted by

Ritika Jonwal

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