Question

Solve this question :- PYQ JEE mains ! ​

Answer 1

Before the cell was connected to an resistor of resistance 10 ohm, the emf (or potential difference of the cell is 4.2 V.

$\implies \sf \: e = 4.2 \: V$

Since,

$\sf \: I = \dfrac{e}{R + r} \\ \\ \implies \sf \: I (10 + r) = 4.2 - - - - (1)$

When cell is connected to a resistor, the terminal potential difference is 4.0 V.

Since,

$\sf \: V = e - I r \\ \\ \implies \sf \: 4 = 4.2 - I r \\ \\ \implies \sf \: I r = 0.2 - - - - - - (2)$

Dividing equations (1) and (2),

$\sf \: \dfrac{r}{10 + r} = \dfrac{0.2}{4.2} \\ \\ \implies \sf \: \dfrac{r}{10 + r} = \dfrac{1}{21} \\ \\ \implies \sf \: 21r = 10 + r \\ \\ \implies \sf \: 20r = 10 \\ \\ \implies \boxed{ \boxed{ \sf r = 0.5 \Omega}}$

Option (D) is correct

Answer 2

Explanation:

As per the provided information in the given question, we have :

Two resistors of 8Ω and 12Ω are connected in parallel and one resistor of 7.2Ω is connected in series with a battery of 6V are connected.

We have been asked to calculate the total resistance.

Let, R₁ be 8Ω ; R₂ be 12Ω and R₃ be 7.2Ω.

Now, as R₁ and R₂ are connected in parallel combination and we know that when resistors are connected in parallel combination then effective resistance is given by,

\begin{gathered}\longrightarrow\boxed{\sf{ \dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} \dots + \dfrac{1}{R_n} }}\\\end{gathered}

⟶

R

P

1

=

R

1

1

+

R

2

1

⋯+

R

n

1

Here,

\begin{gathered}\longrightarrow\sf{ \dfrac{1}{R_{(1,2)}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} }\\\end{gathered}

⟶

R

(1,2)

1

=

R

1

1

+

R

2

1

\begin{gathered}\longrightarrow\sf{ \dfrac{1}{R_{(1,2)}} = \dfrac{1}{8} + \dfrac{1}{12} \; \text{\O}mega}\\\end{gathered}

⟶

R

(1,2)

1

=

8

1

+

12

1

Ømega

\begin{gathered}\longrightarrow\sf{ \dfrac{1}{R_{(1,2)}} = \dfrac{3 + 2}{24} \; \text{\O}mega}\\\end{gathered}

⟶

R

(1,2)

1

=

24

3+2

Ømega

\begin{gathered}\longrightarrow\sf{ \dfrac{1}{R_{(1,2)}} = \dfrac{5}{24} \; \text{\O}mega}\\\end{gathered}

⟶

R

(1,2)

1

=

24

5

Ømega

\begin{gathered}\longrightarrow\sf{ R_{(1,2)} = \dfrac{24}{5} \; \text{\O}mega}\\\end{gathered}

⟶R

(1,2)

=

5

24

Ømega

\begin{gathered}\longrightarrow\underline{\sf{\red{ R_{(1,2)} = 4.8 \; \text{\O}mega}}}\\\end{gathered}

⟶

R

(1,2)

=4.8Ømega

Now, as the combined resistance of R₁ and R₂ are connected in series combination with R₃, so we'll use the formula to calculate resistance in series combination. That is,

\begin{gathered}\longrightarrow\boxed{\sf{ R_S = R_1 + R_2 \dots + R_n}}\\\end{gathered}

⟶

R

S

=R

1

+R

2

⋯+R

n

Here,

\longrightarrow\sf{R_{(1,2,3) }= R_{(1,2)} + R_3}⟶R

(1,2,3)

=R

(1,2)

+R

3

\longrightarrow\sf{R_{(1,2,3) } = (4.8 + 7.2) \; \text{\O}mega}⟶R

(1,2,3)

=(4.8+7.2)Ømega

\begin{gathered}\longrightarrow\underline{\underline{\sf{\red{ R_{(1,2,3)} = 12 \; \text{\O}mega }}}}\\\end{gathered}

⟶

R

(1,2,3)

=12Ømega

Therefore, the total resistance is 12Ω.