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Showing posts with label Electrical circuit. Show all posts
Showing posts with label Electrical circuit. Show all posts
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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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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What must be the emf E in the circuit so that the current flowing through the $7 \Omega$ resistor is $1.80\;A$? Each emf source has negligible internal resistance.
From Figure,
Using Kirchhoff's current law at point $E$
$I_1 \; + \; I_2 = 1.8$ .......... (i)
In closed loop AEFDA,
$24 - E = 3\; I_1 - 2\; I_2$
$24 - E = 3 (1.8 - I_2) - 2 \; I_2$
By Solving,
$E = 5\;I_2 + 18.6$ .......... (ii)
Again, from closed loop EBCFE,
$E = 1.8 * 7 + 2 * I_2$
$E = 12.6 + 2\; I_2$ .......... (iii)
Solving equation (ii) and (iii), we get
$I_2 = - 2\;A$
Then, $I_1 = 3.8\; A$
From equation (iii), we get
$E = 12.6 + 2 * (-2) = 8.6\;V$
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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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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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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 = $I^2*R_2 = (0.091)^2 * 484 = 4\;W$
A cell of internal resistance of 0.2 Ω is connected two cells of resistance 6Ω and 8 Ω joined parallel. There is a current of 0.2 A in the 8 Ω coil. Find the emf of cell.
According to the given information, the electric circuit can be drawn as follows:
Given,
$r = 0.2 \;\Omega$
$R_1 = 8\;\Omega$; $I_1 = 0.2\;A$
$R_2 = 6\;\Omega$
Since $8\;\Omega$ and $6\;\Omega$ are in parallel. So the potential difference across in parallel circuit is equal.
i.e. $I_1\;R_1 = I_2\;R_2$
$I_2 =$$ \frac{I_1\; R_1}{R_2} = \frac{0.2 \; * \; 8}{6}$$ = 0.27 \; A$
Therefore total current $(I) = I_1 + I_2 = 0.2 + 0.27 = 0.47\;A $
External Resistance $(R) = R_1||R_2$$ = \frac{R_1\;R_2}{R_1 + R_2} = $$ \frac{6*8}{6+8}$$= 3.4 \; \Omega$
Now,
ஃ Emf $(E) = IR + Ir = I(R+r) = 0.47(3.4 + 0.2) = 1.7\;V$
Five bulbs are connected in series across 220-Volt line. If one bulb is fused the remaining bulbs are again connected across the same line. Which one of the arrangements will be more illuminated? Justify.
The average power absorbed by each light bulb is directly proportional to the amount of light it emits.
Case I:
Let $R_1 = R_2 = R_3 = R_4 = R_5 = 100\; \Omega$
Then total resistance $R = 500\;\Omega$
Now, Power $(P_1) = $$\frac{V^2}{R} = \frac{(220)^2}{500}$$ = 96.8\;\;Watt$
Case II:
Let $R_1 = R_2 = R_3 = R_4 = 100\; \Omega$
Then total resistance $R = 400\;\Omega$
Now, Power $(P_1) = $$\frac{V^2}{R} = \frac{(220)^2}{400}$$ = 121\;\;Watt$
Since $P_2 > P_1$
Therefore the second case with four light bulbs draws more power and will produce more light.
T6.1 Ohm's Law:
Ohm's Law (V = IR): is a fundamental formula in an electronics. It is used to calculate the relationship between voltage (V), current (I) and resistance (R) in an electrical circuit.
Ohm's Law states that: "The Current through a conductor between two points is directly proportional to the voltage (Potential difference) across the two points" i.e.
$V ∝ I$
$V = I\,R$ .......... (i)
Where $R$ is a proportionality constant called resistance (R) of the conductor. This equation (i) is the mathematical form of Ohm's Law.
An electric circuit is formed when a conductive path is created to allow free electrons to move continuously. This continuous movement of free electrons through the conductor of a circuit is called a Current (I).
The force motivating electrons to flow in a circuit is called Voltage (V). The voltage is a specific measure of potential energy that is always relative between two points.
Free electrons tends to move through conductors with some degree of friction (or, opposition to motion), called Resistance (R).
The amount of current in a circuit depends on the amount of voltage available to motivate the electrons, and the amount of resistance in the circuit to oppose electron flow (Ohm's Law).
Ohm's Law Calculator: Click here.
Experimental Verification of Ohm's Law:
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