Current Electricity - NEET Physics Questions
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Current Electricity

Question 21: difficult

For given circuit, heat produced by a current in resistance of 5Ω is 10 Cal/sec. Then the heat produced in resistance of 4Ω is

1. 2 Cal / sec
2. 2.5 Cal / sec
3. 4 Cal / sec
4. 5 Cal / sec
View Answer
Question 22: moderate

Some light bulbs are conected in parallel to a 120 V source as shown in the figure. Each bulb dissipates an average power of 60 W.The circuit has a fuse F that burns out when the current in the circuit exceeds 9 A. Determine the largest number of bulbs of the following, that can be used in the circuit without burning out the fuse.

1. 9
2. 17
3. 36
4. 34
View Answer
Question 23: moderate

Four identical bulbs each rated 100 watts, 220 volts are connected across a battery as shown. The power consumed by them is:

1. 75 watt
2. 400 watt
3. 300 watt
4. 400/3 watt
View Answer
Question 24: moderate

Six identical light bulbs are connected to a battery to form the circuit shown. Which light bulb(s) glow the brightest?

1. 1, 2 and 3
2. 5 and 6
3. 1, 2, 3 and 4
4. 4 only
View Answer

Current through bulb 4 is maximum

Question 25: easy

A carbon resistor is marked with the rings coloured brown, black green and gold. The resistance (in ohm) is :

1. \[3.2\times 10^{5}\] ±5%
2. \[1\times 10^{6}\] ±10%
3. \[1\times 10^{7}\] ±5%
4. \[1\times 10^{6}\] ±5%
View Answer

To determine the resistance of a carbon resistor with color bands brown, black, green, and gold, use the color code for resistors:


1. Color code values:

  • Brown:
    11
     

    (1st digit)

  • Black:
    00
     

    (2nd digit)

  • Green:
    10510^5
     

    (multiplier)

  • Gold:
    ±5%\pm 5\%
     

    (tolerance)


2. Calculate resistance:

The resistance is calculated as:

 

R=(1st digit×10+2nd digit)×multiplier.R = (\text{1st digit}\,\times\,10 + \text{2nd digit}) \,\times\, \text{multiplier}.

 

Substitute values:

 

R=(1×10+0)×105=10×105=106Ω.R = (1 \times 10 + 0) \times 10^5 = 10 \times 10^5 = 10^6 \, \Omega.

 

This is equal to:

 

R=1MΩ(megaohm).R = 1 \, \text{M}\Omega \, (\text{megaohm}).

 


3. Tolerance:

The gold band indicates a tolerance of

±5%\pm 5\%

, so the resistance can vary between:

 

1MΩ±5%.1 \, \text{M}\Omega \pm 5\%.

 


Final Answer:

The resistance is:

 

1MΩ±5%.\boxed{1 \, \text{M}\Omega \, \pm 5\%.}

 

Question 26: moderate

N identical cells, each of emf e and internal resistance r, are joined in series. Out of these, n cells are wrongly connected, i.e., their terminals are connected in reverse of that required for series connection. n < N/2. Let  \(\varepsilon_{0}\) be the emf of the resulting battery and \( r_{0}\) be its internal resistance,

1. \[\varepsilon_{0}=\left( N-n \right)\varepsilon, r_{0}=\left( N-n \right)r\]
2. \[\varepsilon_{0}=\left( N-2n \right)\varepsilon, r_{0}=\left( N-2n \right)r\]
3. \[\varepsilon_{0}=\left( N-2n \right)\varepsilon, r_{0}=Nr\]
4. \[\varepsilon_{0}=\left( N-n \right)\varepsilon, r_{0}=Nr\]
View Answer

When

NN

identical cells, each of emf

ee

and internal resistance

rr

, are connected in series and

nn

cells are connected in reverse, the resulting emf and internal resistance of the battery can be determined as follows:


1. Resultant emf ( ε0\varepsilon_0

 

):

  • For correctly connected cells, the total emf is:
    εcorrect=(Nn)e.\varepsilon_{\text{correct}} = (N - n)e.
     
  • For n reversed cells, their emf opposes the total emf. The opposing emf is:
    εreverse=ne.\varepsilon_{\text{reverse}} = ne.
     

    The net emf of the resulting battery is: 

    ε0=εcorrectεreverse=(Nn)ene=(N2n)e.\varepsilon_0 = \varepsilon_{\text{correct}} - \varepsilon_{\text{reverse}} = (N - n)e - ne = (N - 2n)e. 


2. Resultant internal resistance ( r0

 

):

  • All cells, whether correctly or incorrectly connected, contribute to the total internal resistance because resistances add in series. The total internal resistance is:
    r0=Nr.r_0 = N r.
     

Final Answer:

The emf and internal resistance of the resulting battery are:

ε0=(N2n)e,r0=Nr.\boxed{\varepsilon_0 = (N - 2n)e, \quad r_0 = Nr.}

 

Question 27: moderate

The ratio of the resistance of conductor at temperature 15°C to its resistance at temperature 37.5°C is 4 : 5. the temperature coefficient of resistance of the conductor is : (reference is taken as 0°C)

1. \[1/25 ^{o}C^{-1}\]
2. \[1/50 ^{o}C^{-1}\]
3. \[1/80 ^{o}C^{-1}\]
4. \[1/75 ^{o}C^{-1}\]
View Answer

To find the temperature coefficient of resistance

α\alpha

of the conductor, we use the formula for the change in resistance with temperature:

 

RT=R0(1+αT)R_T = R_0 \left( 1 + \alpha T \right)

 

Where:


  • RTR_T
     

    is the resistance at temperature TT 

    ,


  • R0R_0
     

    is the resistance at the reference temperature (0°C),


  • α\alpha
     

    is the temperature coefficient of resistance,


  • TT
     

    is the temperature change in °C.

Step 1: Given

  • The ratio of resistances at 15°C and 37.5°C is given as
    R15R37.5=45\frac{R_{15}}{R_{37.5}} = \frac{4}{5}
     

    .

  • The resistance at temperature 15°C,
    R15=R0(1+α×15)R_{15} = R_0 (1 + \alpha \times 15)
     

    .

  • The resistance at temperature 37.5°C,
    R37.5=R0(1+α×37.5)R_{37.5} = R_0 (1 + \alpha \times 37.5)
     

    .

Step 2: Set up the equation based on the given ratio

 

R15R37.5=45\frac{R_{15}}{R_{37.5}} = \frac{4}{5}

 

Substituting the expressions for

R15R_{15}

and

R37.5R_{37.5}

:

 

R0(1+α×15)R0(1+α×37.5)=45\frac{R_0 (1 + \alpha \times 15)}{R_0 (1 + \alpha \times 37.5)} = \frac{4}{5}

 

Canceling

R0R_0

from both the numerator and denominator:

 

1+15α1+37.5α=45\frac{1 + 15\alpha}{1 + 37.5\alpha} = \frac{4}{5}

 

Step 3: Solve for α\alpha

 

Cross-multiply to solve for

α\alpha

:

 

5(1+15α)=4(1+37.5α)5(1 + 15\alpha) = 4(1 + 37.5\alpha)

 

Expanding both sides:

 

5+75α=4+150α5 + 75\alpha = 4 + 150\alpha

 

Simplify:

 

54=150α75α5 - 4 = 150\alpha - 75\alpha

 

1=75α1 = 75\alpha

 

α=175\alpha = \frac{1}{75}

 

Final Answer:

The temperature coefficient of resistance

α\alpha

is

175per °C\boxed{\frac{1}{75}} \, \text{per °C}

.

Question 28: moderate

The equivalent resistance of the network shown in the figure between the points A and B is :-

 

1. 6 Ω
2. 8 Ω
3. 16 Ω
4. 24 Ω
View Answer
Question 29: moderate

The effective resistance across the points A and I is :-

1. 2 Ω
2. 1 Ω
3. 0.5 Ω
4. 5 Ω
View Answer
Question 30: moderate

The resultant resistance value of n resistance each of r ohm when connected in parallel, is x. When these n resistances are connected in series, the resultant value is :

 

1. nx
2. n²x
3. rnx
4. r²x/n
View Answer

Let's break this down step by step.

1. Resistances connected in parallel

When

nn

resistors, each of resistance

rr

, are connected in parallel, the total or equivalent resistance

RparallelR_{\text{parallel}}

is given by the formula:

 

1Rparallel=1r+1r++1r(n terms)\frac{1}{R_{\text{parallel}}} = \frac{1}{r} + \frac{1}{r} + \cdots + \frac{1}{r} \quad \text{(n terms)}

 

This simplifies to:

 

1Rparallel=nr\frac{1}{R_{\text{parallel}}} = \frac{n}{r}

 

So, the equivalent resistance is:

 

Rparallel=rnR_{\text{parallel}} = \frac{r}{n}

 

We're told that the resultant resistance when connected in parallel is

xx

, so:

 

x=rnx = \frac{r}{n}

 

2. Resistances connected in series

When the same

nn

resistors, each of resistance

rr

, are connected in series, the total or equivalent resistance

RseriesR_{\text{series}}

is simply the sum of the individual resistances:

 

Rseries=r+r++r(n terms)R_{\text{series}} = r + r + \cdots + r \quad \text{(n terms)}

 

So:

 

Rseries=n×rR_{\text{series}} = n \times r

 

3. Relating the two scenarios

From the parallel connection, we know that:

 

x=rnx = \frac{r}{n}

 

Thus, the resistance when the resistors are connected in series is:

 

Rseries=n×r=n×(x×n)=r×n×xR_{\text{series}} = n \times r = n \times \left( x \times n \right) = r \times n \times x

 

Therefore, the resistance when these

nn

resistors are connected in series is:

 

r×n×x\boxed{r \times n \times x}