Chapter - Electrical Circuit | Part 3

 

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The driver cell of a potentiometer has an e.m.f. of 2 V and negligible internal resistance. The potentiometer wire has a resistance of 3 Ω. Calculate the resistance needed in series with the wire if a p.d. of 1.5 mV is required across the whole wire.

Given,

Emf $(E) = 2\;V$

Resistance of Potentiometer wire $(R) = 3\; \Omega$

Potential Difference $(V) = 1.5 \; mV = 1.5 * 10^{-3}\;V$

Let, $R'$ be the resistance to be connected in series.

$\therefore$ Total resistance of the circuit $(R_T) = R' + 3$

Here, current across the potentiometer circuit $(I) = \frac{E}{R_T} = \frac{2}{R'\; + \; 3}$

Also, we have, p.d. across potentiometer wire $ = IR$

or, $1.5 * 10^{-3} = \frac{2}{R' \; + \; 3} * 3$

or, $R' + 3 = 4000$

$\therefore$ $ R' = 3997 \; \Omega$

Hence resistance needed in series $= 3997 \; \Omega$

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The e.m.f of a battery A is balanced by a length 75 cm on a potentiometer wire. The emf of a standard cell 1.02 V is balanced by a length of 50 cm. What is the emf of A?

Given,

Let $E_1$ be the emf of a cell A

Balancing length for $A \; (l_1) = 75\;cm$

Emf of standard cell $(E_2) = 1.02\;V$

Balancing length for $E_2 \; (l_2) = 50\;cm $

We know,

$\frac{E_1}{E_2} = \frac{l_1}{l_2}$

$E_1 = \frac{75}{50} * 1.53\;V$

Hence, emf of the cell A is $11.53\;V$

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The driver cell of a potentiometer has an e.m.f. of 2 V and negligible internal resistance. The potentiometer wire has a resistance of 3 Ω. Calculate the resistance needed in series with the wire if a pd 5 mV is required across the whole wire. The wire is 100 cm long and a balanced length of 60 cm is obtained for a thermocouple of emf E. What is the value of E?

Consider $AB$ is the potentiometer wire,

$V = 2\;V$

$R_{AB} = 3 \; \Omega$ 

Let $R$ be the resistance needed in the series, the the potential difference across $AB = 5\;mV = 5 * 10^{-3}\;V$

i.e. $I\;R_{AB} = 5\;mV$

or, $\frac{V}{R_{AB} + R} * R_{AB} = 5 * 10^{-3}$

or, $\frac{2}{3 \; + \; R} * 3 = 5 * 10^{-3}$

By Solving, we get

$R = 1197 \Omega$

Again,

$l_1 = 100 \; cm $

$V_1 = 5\; mV $

$l_2 = 60 \; cm$

$V_2 = ?$

We have,

$\frac{V_1}{V_2} = \frac{l_1}{l_2}$

$\frac{5}{V_2} = \frac{100}{60}$

$\therefore$  $V_2 = 3\;mV$

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Using Kirchhoff's laws of current and voltage, find the current in $2 \; \Omega$ resistor in the given circuit.

Given,

Applying Kirchhoff's current law at point $'B'$.

$I_3 = I_1  + I_2$ .......... (i)

Now, Applying Kirchhoff's voltage law in close loop ABEFA

$35 = I_1 * 3 + I_3 * 2$

$35 = 3 \; I_1 + 2 * (I_1 + I_2)$

or, $3\;I_1 + 2 \; I_2 + 2 \; I_2 = 35$

or, $5\; I_1 + 2 \; I_2 = 35$ ......... (ii)

Again, Applying Kirchhoff's Voltage law in close loop CBEDC,

$40 = I_2 * 4 + I_3 * 2$

or, $40 = I_2 * 4 + I_3 * 2$

or, $40 = 4\; I_2 + 2 (I_1 + I_2)$

or, $2\; I_1 + 6 \; I_2 = 40$ .......... (iii)

Solving equation (ii) and (iii) we get,

$I_1 = 5\;A$       and       $I_2 = 5\; A$

So current through $2 \;\Omega = I_3 = I_1 + I_2 = 5 + 5 = 10 \; A$

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The total length of the wire of a potentiometer is 10 m. A potential gradient of 0.0015 V/cm is obtained when a steady current is passed through this wire . Calculate: a) The distance of the null point on connecting standard cell of 1.018 V. b) The unknown p.d. if the null point is obtained at a distance of 940 cm c) The maximum p.d. which can be measured by this instrument.

Given,

The length of the potentiometer $(L) = 10\;m$

Potential gradient $(K) = 0.0015\; V/cm = 0.15 \; V/m$

a) Distance of null point $(l_1) = ?$

Voltage of the cell $(V) = 1.018 \; V$

From the principle of the potentiometer, We have,

$V \propto l$

or, $V = k\;l_1$

or, $1.018 = 0.15 * l_1$

or, $l_1 = 6.8 \; m$

b) Potential Difference $(V) = ?$

Length of null point $(l_2) = 940 \; cm = 9.4 \; m$

Again from the principle of the potentiometer,

$V \propto l_2$

or, $V = k * l_2 = 0.15 * 9.4 = 1.41\;V$

c) Maximum P.d. $(V_{max}) = ?$

$V_{max} = k * L = 0.15 * 10 = 1.5 \; V$

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A simple potentiometer circuit is a setup as in figure, using uniform wire AB, 1m long, which has a resistance of 2 Ω. The resistance of the 4 V battery is negligible. If the variable resistor R were given a value of 2.4 Ω, what would be the length AC for zero galvanometer deflection?

Given,

Length of wire $AB$ $(l_{AB}) = 1\;m$

Resistannce of wire $(R) = 2.4 \; \Omega$

EMF of cell $(E_1) = 4\; V$

Resistance across $AB$ $(R_{AB}) = 2 \; \Omega$

Length of $AC$ $(l_{AC}) = ?$

For zero galvanometer deflection,

P.D. across $AC = 1.5 \; V$

And, $V = I * R_{AC}$

or, $1.5 = \frac{4}{2.4 \; + \; 2} * R_{AC}$

$R_{AC} = \frac{1.5 \; * \; 4.4}{4}$  $= 1.65 \; \Omega$

Since, $2 \; \Omega$ wire $AB$ has length $1\;m$

$1 \; \Omega$ wire has length $\frac{1}{2}\;m$

$\therefore$ $1.65 \Omega$ wire $AB$ has length $\frac{1}{2} * 1.65 = 0.825 \; m$

$\therefore$ Length of $AC = 0.825 \; m.$

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A potentiometer is 10 m long. It has a resistance of 20 Ω. It is connected in series with a battery of 3 V and a resistance of 10 Ω. What is the potential gradient along with wire?

Given,

Length of potentiometer wire $(l) = 10\;m$

Resistance of potentiometer $(R_p) = 20\; Ω$

Potential of battery $(V_b) = 3V$

Resistance of battery $(R_b) = 10 \; Ω$

Potential Gradient $(\frac{V}{l}) = \;?$

Let $I$ be the current flowing through the circuit.

Then, from Ohm's law,

$V_b = IR$     [where R= Total resistance of the circuit = $R_b + R_p$

or, $3 = I\;*\;(10+20)$

⇒ I = 0.1 A

Since the combination is in series, the current (I) all over the circuit is same.

So, p.d in AB $(V_{AB})$ = I * $R_{b}$ = 0.1 * 20 = 2 V

Now, Potential Gradient $(\frac{V}{l}) = \frac{2}{10}$ = 0.2V/m

Hence, the potential gradient along the wire is 0.2 V/m.

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