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 battery of $6 \; V$ and internal resistance $0.5\; \Omega$ is joined in parallel with another of $10\;V$ and internal resistance $1\;\Omega$. The combination sends a current through an external resistance of $12\; \Omega$. Find the current through each battery.

Given,

Emf of one cell $(E_1) = 6\; V$

Emf of another cell $(E_2) = 10\; V$

Internal resistance of one cell $(r_1) = 0.5 \; \Omega$

Internal resistance of another cell $(r_2) = 1 \; \Omega$

External resistance $(R) = 12 \; \Omega$

Current through $E_1 \; (I_1) = ?$

Current through $E_2 \; (I_2) = ?$

Using Kirchhoff's voltage law in closed-loop ABXCDA, we have

$E_1 = I_1\; r_1 + I_3 \; R$

$6 = I_1\;r_1 + (I_1 + I_2)\;R$

or, $6 = 0.5\;I_1 + 12\;I_1  + 12 \; I_2$

or, $6 = 12.5 \; I_1 + 12 \; I_2$ .......... (i)

Again, Using Kirchhoff's voltage law in closed loop ABYCDA, we have

$E_2 = I_2\; r_2 + I_3 \; R$

or, $10 = I_2 \; r_2 + (I_1 + I_2)\;R$

or, $10 = 12 \; I_1 + 13\; I_2$ .......... (ii)

By Solving equation (i) and (ii) we get

$I_1 = -2.27 \; A$       and       $I_2 = 2.86 \; A$

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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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In the adjacent circuit find, (i) the current in resistor $R$, (ii) the unknown emf $E$, (iii) the resistance $R$, (iv) if the circuit is broken at $P$, what is the current in resistor $R$?

(i) Using Kirchhoff's current law at point $'D'$, 

Current through $R = 6 - 4 = 2\;A$ 

(ii) Applying Kirchhoff's voltage law in close loop ABCDA, 

$E = 3 \; *\; 6 \;+\; 4\; *\; 6 = 18\; + \;24 = 42\; V$ 

(iii) Again, Applying Kirchhoff's voltage law in close loop AFEDA, 

$E - 28 = 6 * 4 - R * 2$ 

$42 - 28 = 24 - 2R$ 

$2R = 24 - 14 = 10$

$R = 5 \; \Omega$

(iv) If the circuit is broken from point $'P'$ then,

Current$(I) = \frac{V}{R\; +\; 3} = \frac{28}{5\; + \;3}$   $ = 3.5\; A$

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Two lamps rated 25 W - 220 V and 100 W - 220 V are connected to 220 V supply. Calculate the powers consumed by the lamps.

Given, 

First lamp = 25 W - 220 V 

Second lamp = 100 W - 220 V 

Voltage (V) = 220 V 

Power Consumed (P) = ?

Now, to calculate the resistance

For first lamp,   $P_1 = 25\;W$ 

Voltage $(V_1) = 220\;V$ 

$R_1 = $$ \frac{V_1^{2}}{P_1} = \frac{220^2}{25}$$ = 1936\; \Omega$                                                   [∵ $P = \frac{V^2}{R}$]

For second lamp,   $P_2 = 100\;W$ 

Voltage $(V_2) = 220\;V$ 

$R_2 = $$\frac{V_2^{2}}{P_2} = \frac{220^2}{100}$$ = 484 \; \Omega$ 

If two lamps are connected in series and joined to 220 V mains, then the current  in the circuit (I) is calculated as,

$I = $$\frac{V}{R_1 + R_2} = \frac{220}{1936 + 484}$$ = 0.091 \;A$

Finally, 

Power consumed by the first lamp = $I^2*R_1 = (0.091)^2 * 1936 = 16\;W$ 

Power consumed by the second lamp = $I^2*R_2 = (0.091)^2 * 484 = 4\;W$

T6.1 Electrical Resistance and Resistivity:

» Resistance (Symbol R): 

Its ability to oppose the flow of charge through it. In general, the resistance is the ratio of the potential difference across an electrical component to the current passing it.

Mathematically,

$R = \frac{V}{I}$ .......... (i)

Where, V = Voltage and I = Current. It's unit is ohm $(\Omega)$ in SI unit.

The one ohm $(1 \; \Omega)$ of the conductor is defined as, if one ampere current flows through the resistance under a potential difference of one volt. i.e.

$1\; \Omega = \frac{1\; Volt}{1\; Ampere}$ .......... (ii)

Click here for the Resistance Colors code

» Resistivity (Symbol $\rho$): 

The electrical resistivity of a conductor (material) is a measure of how strongly the material opposes the flow of electric current through it. The resistivity is depends upon the lengths and cross-sectional areas of the conductor. The higher the resistivity $\rho$ the more the resistance and vice-versa.

For example: The resistivity of a copper (good conductor) is in the order of 1.72 * 10$^-8$ Ω m. Whereas the resistivity of a air (insulator / poor conductor) 10$^{10}$ 10 $^{14}$ Ω m. 

The resistivity of a conductor s defined as the resistance of the conductor of unit cross-sectional area per unit length. It's unit is ohm meter (Ω m), in SI system.

$\rho = R\frac{A}{l}$ .......... (iii)

 

Mathematically, 

Let length $l$ of the conductor having resistance $R$ and its cross sectional area $A$.

At constant temperature,

The resistance $R$ of the conductor is directly proportional to the length $l$ of the conductor.

$R ∝ l$ .......... (iv)

And, the resistance $R$ of the conductor is inversely proportional to the cross sectional area $A$.

$R ∝ \frac{1}{A}$ .......... (v)

From equation (iv) and (v), we get

$R ∝ \frac{l}{A}$

or, $R = \rho \frac{l}{A}$ .......... (vi)

Where $\rho$ is a proportionality constant called 'Resistivity of a conductor'.

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