SATHEE: current-electricity Question 42

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Question: Q. 4. A uniform wire of resistance $48 Ω$ is cut into three pieces so that the ratio of the resistances $R_{1} : R_{2}$ : $R_{3} = 1 : 2 : 3$ and the three pieces are connected to form a triangle across which a cell of emf $8 V$ and internal resistance $1 Ω$ is connected as shown. Calculate the current through each part of the circuit.

A [O.D. Comptt. I, II, III 2013]

$8 V$

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Solution:

Ans.

$R_{1} : R_{2} : R_{3} = 1 : 2 : 3$

Let,

$\begin{aligned} R_{1} = R & R_{2} = 2 R & R_{3} = 3 R \end{aligned}$

Total resistance $(R_{T}) = R + 2 R + 3 R$

$\begin{array}{llrl} = 48 Ω ∴ & R & = 8 Ω Hence & R_{1} & = 8 Ω & R_{2} & = 16 Ω and & R_{3} & = 24 Ω \end{array}$

According to circuit :

$\begin{aligned} E = I (R_{T} + r) & 8 = I (R_{T} + r) & 8 = I (12 + 1) & I = \frac{8}{13} A & I = 0.615 A & 8 \end{aligned}$

Let current across $R_{1}$ is $I_{1}$

and current across $R_{2}$ is $I_{2}$

So current across

$I_{2} = I_{1}$

$R_{3} = I_{3} = (I - I_{1})$

Apply Kirchhoff’s $2^{nd}$ Law

$\begin{aligned} I_{1} & = \frac{R_{3}}{(R_{1} + R_{2} + R_{3})} \times I & = \frac{24}{(8 + 16 + 24)} \times 0.615 1 / 2 I_{2} & = I_{1} = 0.3075 A I_{3} & = I - I_{1} & = 0.615 - 0.3075 & = 0.3075 A \end{aligned}$

[AT Q. 5. The network $P Q R S$ , shown in the circuit diagram, has the batteries of $4 V$ and $5 V$ and negligible internal resistance. A milliammeter of $20 Ω$ resistance is connected between $P$ and $R$ . Calculate the reading in the milliammeter.

A [O.D. Comptt. I, II, III 2012]

Ans. Let us redraw circuit as shown.

Using Kirchhoff’s voltage law in closed loop $D A B C D$

$\begin{matrix} (i) & (I_{1} + I_{2}) R_{3} + I_{1} R_{1} - E_{1} = 0 \end{matrix}$

In a closed loop $A B F E A$

$\begin{matrix} (ii) & (I_{1} + I_{2}) R_{3} + I_{2} R_{2} - E_{2} = 0 \end{matrix}$

Substituting the values of $Misplaced &$ in equations (i) & (ii)

We have $44 I_{1} + 4 I_{2} = 1$

$and 5 I_{1} + 20 I_{2} = 1$

Solving above equations, we get

$Misplaced &$

Now, current in milliammeter

$\begin{aligned} I_{m} & = I_{1} + I_{2} & = 0 \cdot 0186 A + 0 \cdot 0454 A & = 0 \cdot 064 A = 64 mA \end{aligned}$

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एनसीईआरटी पुस्तकें और समाधान

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