Combination of Batteries - NEET Physics Questions
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Combination of Batteries

Question 1: 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 2: easy

Two cells of e.m.fs. E1 and E2 and internal resistance r1 and r2 are connected in parallel. Then the e.m.f. and internal resistance of the equivalent source is :

1. \[E_{1}+E_{2} and \frac{r_{1}r_{2}}{r_{1}+r_{2}}\]
2. \[E_{1}-E_{2} and r_{1}+r_{2}\]
3. \[\frac{E_{1}r_{2}+E_{2}r_{1}}{r_{1}+r_{2}} and \frac{r_{1}r_{2}}{r_{1}+r_{2}}\]
4. \[\frac{E_{1}r_{2}+E_{2}r_{1}}{r_{1}+r_{2}} and r_{1} +r_{2}\]
View Answer

To find the equivalent emf (

EeqE_{\text{eq}}

) and internal resistance (

reqr_{\text{eq}}

) of two cells connected in parallel, we use the following principles:


1. Equivalent emf ( EeqE_{\text{eq}}

 

):

In parallel connection, the total current is the sum of the currents through each cell. Using Kirchhoff's Voltage Law, the equivalent emf is given by:

 

Eeq=E1r2+E2r1r1+r2.E_{\text{eq}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}.

 


2. Equivalent internal resistance ( reqr_{\text{eq}}

 

):

For resistances in parallel, the equivalent resistance is given by:

 

1req=1r1+1r2.\frac{1}{r_{\text{eq}}} = \frac{1}{r_1} + \frac{1}{r_2}.

 

Simplify:

 

req=r1r2r1+r2.r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2}.

 


Final Answer:

The equivalent emf and internal resistance of the parallel combination are:

 

Eeq=E1r2+E2r1r1+r2,req=r1r2r1+r2.\boxed{E_{\text{eq}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}, \quad r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2}}.

 

Question 3: easy

When a resistance of 2 ohm is connected across the terminals of a cell, the current is 0.5 amp. When the resistance is increased to 5 ohm, the current is 0.25 amp. The emf of the cell is:

1. 1.0 volt
2. 2.0 volt
3. 1.5 volt
4. 2.5 volt
View Answer

Using \( E = I(R + r) \), we set up equations: \( E = 0.5(2 + r) \) and \( E = 0.25(5 + r) \). Equating them gives \( r = 1\ \Omega \), which yields \( E = 0.5(2 + 1) = 1.5\text{ V} \).