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Class 10 Class 12

(i) State the principle of working of a meter bridge.
(ii) In a meter bridge balance point is found at a distance l₁ with resistors R and S as shown in the figure.

When an unknown resistor X is connected in parallel with the resistor S, the balance point shift to a distance l₂. Find the expression for X in terms of l₁ l₂ and S.

(i) A slide wire bridge is known as meter bridge. It is constructed on the principle of balanced Wheatstone bridge, when a Wheatstone bridge is balanced then

straight P over straight Q space equals space fraction numerator straight l over denominator 100 minus straight l end fraction

(ii) When resistors R and S are connected. Since balance point is found at a distance l₁from the zero and

therefore

straight R over straight S space equals space fraction numerator straight l subscript 1 over denominator 100 minus straight l subscript 1 end fraction

...(i)
When unknown resistance X is connected in parallel to S.
∴ total resistance in the right hand gap is

straight S subscript 1 space equals space fraction numerator SX over denominator straight S plus straight X end fraction

open square brackets because 1 over straight R space equals space 1 over straight R subscript 1 plus 1 over straight R subscript 2 space space rightwards double arrow space space straight R space equals space fraction numerator straight R subscript 1 straight R subscript 2 over denominator straight R subscript 1 plus straight R subscript 2 end fraction close square brackets

Since the balance point is obtained at a distance l₂ from the zero end

therefore

straight R over straight S subscript 1 space equals space fraction numerator straight l subscript 2 over denominator 100 minus straight l subscript 2 end fraction

putting the value of S₁, we get

fraction numerator straight R over denominator begin display style fraction numerator SX over denominator straight S plus straight X end fraction end style end fraction space equals space fraction numerator straight l subscript 2 over denominator 100 minus straight l subscript 2 end fraction fraction numerator straight R space left parenthesis straight S plus straight X right parenthesis over denominator SX end fraction space equals space fraction numerator straight l subscript 2 over denominator 100 minus straight l subscript 2 end fraction space space space space space space space space space space space space space space space space space space space space... left parenthesis ii right parenthesis

Dividing equation (ii) by (i), we get
(i) A slide wire bridge is known as meter bridge. It is constructed o

4240 Views

State the principle of potentiometer. Draw a circuit diagram used to compare the emf of two primary cells. Write the formula used. How can the sensitivity of a potentiometer be increased?

Principle of potentiometer. The fall of potential along any length of the wire is directly proportional to that length. When a constant current flows through a wire of uniform cross-section and composition.
V ∝ l
Comparison of emfs of two primary cells: The circuit diagram is shown in the figure.

When the key K is closed, a constant current flows the potentiometer wire. By closing key K₁, the cell E₁ is included in the circuit. The jockey is adjusted till galvanometer shows no deflection. Suppose AJ₁ = l₁ is the balancing length for cell E₁. Then
E₁ = kl₁where k is the potential gradient. Now the null point is obtained for cell E₂ by closing key K₂. Let AJ₂ = l₂ be the balancing length in this case. Then E₂ = kl_{2
$therefore space space space space space box enclose straight E subscript 2 over straight E subscript 1 space equals space straight l subscript 2 over straight l subscript 1 end enclose$}by increasing the length of a potentiometer's wire the sensitivity of a potentiometer can be increased.

4120 Views

Two cells of emf 1.5 V and 2 V and internal resistance 1 ohm and 2 ohm respectively are connected in parallel to pass a current in the same direction through an external resistance of 5 ohm.
(a) Draw the circuit diagram.
(b) Using Kirchhoff’s laws, calculate the current through each branch of the circuit and potential difference across the 5 ohm resistor.

(i) The circuit diagram is shown below:

(ii) (a) Let I₁, I₂ and I₁+ I₂ be the currents flowing through the resistors r₁, r₂ and R respectively. Applying Kirchhoff’s law to the closed circuit CBAFC, we have
– 2 I₂ + 1 I₁ = 2.0 – 1.5 = 0
or, 2 I₂ – I₁ = 0.5 ...(i)
Again applying Kirchhoff’s law for closed circuit CFEDC, we have
– 1 I₁ – 5 (I₁ + I₂) + 1.5 = 0
or, 6 I₁ + 5 I₂ = 1.5 solving (i) and (ii), we get
$straight I subscript 2 space equals space 45 over 17 space equals space 9 over 34 straight A$
$therefore$ $straight I subscript 1 space equals space 5 over 170 straight A$
Current through CF = $straight I subscript 1 space equals space 5 over 170 straight A$

Current through BA = $straight I subscript 2 space equals space 9 over 34 straight A$
$therefore$ Current through DE = $straight I subscript 1 plus straight I subscript 2 space equals space 5 over 170 plus 9 over 34$
$equals space 150 over 170 straight A space equals space 5 over 17 straight A$
(b) Potential difference across 5 Ω resistor
$equals space left parenthesis straight I subscript 1 plus straight I subscript 2 right parenthesis space cross times 5 equals space 5 over 17 cross times 5 space equals space 1.47 space straight V.$

6336 Views

In a power station, a copper bar designed to carry many amperes of current is 2 m long and 10 cm² in cross-section. Determine the resistance of the bar at 0 °C. What potential difference is needed to cause a current of 5000 A through the bar? The resistivity of copper at 0°C is 1.59 x 10^–8 Ω m. Also compute the resistance of the bar if it is stretched to form a long and uniform wire of 1 mm² cross-section.

Length of the copper bar, l = 2m
Area of cross-section of the bar, A=10 cm²= 10 = 10³ m²
Current across the bar, I = 5000 A
Resistivity of Copper at 0^o C,

ρ =

1.59 x 10^–8 Ω m
The resistance of the bar is given by,

R = \frac{ρl}{A} = \frac{1.59 \times 10^{- 8} \times 2}{10^{- 2}} = 3.18 \times 10^{- 5} Ω

Now, using Ohm's law, the potential difference across its ends is

V = I R = 5000 \times 3.18 \times 10^{- 5} = 0.159 V

Volume of the bar = A.l = 2 x 10^–3 m³If the bar is stretched to form a long and uniform wire then,

Area of cross-section of the wire = 1 mm² = 10^–6 m²Length of wire, l

\frac{volume}{area} = \frac{2 \times 10^{- 3}}{1 \times 10^{- 6}} = 2 \times 10^{3} m

The wire has the same amount of copper as the bar, but it is 2000 m long.
Therefore, the resistance of the wire is

R = \frac{ρl}{A} = \frac{1.59 \times 10^{- 8} \times 2000}{10^{- 6}} = 31.8 Ω

188 Views

We have 30 watt, 6 volt bulb which we want to glow by a supply of 120 V. What will have to be done for it?

Given,
Power of the bulb, P = 30 W
Potential of the bulb, V = 6 V
Supply voltage, V' = 120 V

Now, Resistance of the bulb,

R = \frac{V^{2}}{P} = \frac{(6)^{2}}{30} = 1.2 Ω

Current that can be flown across the bulb,

I = \frac{P}{V} = \frac{30}{6} = 5 A

Let R’ be the resistance used in series with the bulb to have a current of 5 A in the circuit.

∴

Total resistance = R' + R = (R' + 1.2)

And so,
Current, I = V'/(R'+1.2)

5 = \frac{120}{R' + 1.2}

R' = \frac{120}{5} - 1.2 = 22.8 Ω

That is, a resistance of 22.8

Ω

will have to be used in series with the lamp so as to supply a voltage of 120 V.

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Current Electricity

State the principle of potentiometer. Draw a circuit diagram used to compare the emf of two primary cells. Write the formula used. How can the sensitivity of a potentiometer be increased?