Physics Current Electricity

The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of 15Ω resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10V is maintained across AC.

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Contributor-Level 10

Let V? = 10V, V? = xV, V? = 0V, and V? = yV.
Applying Kirchhoff's current law at node B:
(x - 10)/100 + (x - y)/15 + (x - 0)/10 = 0 ⇒ 53x - 20y = 30 . (1)
Applying Kirchhoff's current law at node D:
(y - 10)/60 + (y - x)/15 + (y - 0)/5 = 0 ⇒ 17y - 4x = 10 . (2)
Solving equations (1) and (2), we get:
x = 0.865 and y = 0.792
The current i? is:
i? = (x - y) / 15 = 4.87 mA

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In the experiment of Ohm's law, a potential difference of 5.0V is applied across the end of a conductor of length 10.0 cm and diameter of 5.00 mm. The measured current in the conductor is 2.00 A. The maximum permissible percentage error in the resistivity of the conductor is:

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Raj Pandey

Contributor-Level 9

From ohm's Law, V = IR = I (ρl / A) = I (ρl / (πd²/4) ⇒ ρ = (πd²V) / (4lI)
Relative error in resistivity,
Δρ/ρ = 2 (Δd/d) + ΔV/V + Δl/l + ΔI/I = 2 * (0.01/5.00) + (0.1/5.0) + (0.1/10.0) + (0.01/2000) = 0.039
Percentage error = (Δρ/ρ) * 100 = 3.9%

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The equivalent resistance of series combination of two resistors is 's'. When they are connected in parallel, the equivalent resistance is 'p'. If s = np, then the minimum value for n is . (Round off to the Nearest Integer)

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For series combination: s = R? + R?

For parallel combination: p = (R? ) / (R? + R? )

Given the condition s = np:
R? + R? = n * (R? ) / (R? + R? )
(R? + R? )² = nR? R?
R? ² + 2R? R? + R? ² = nR? R?
R? ² - 2R? R? + R? ² + 4R? R? = nR? R?
(R? - R? )² = (n - 4)R?
(R? - R? )² / (R? ) = n - 4
n = 4 + (R? - R? )² / (R? )

Since (R? - R? )² is always non-negative, the minimum value of the term (R? - R? )² / (R? ) is 0. This occurs when R? = R?
Therefore, the minimum value of n is 4.

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A current of 10 A exists in a wire of cross-sectional area of 5mm² with a drift velocity of 2*10⁻³ ms⁻¹. The number of free electrons in each cubic meter of the wire is ……………

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Given:

i = 10
e = 1.6 x 10? ¹? C
A = 5 x 10?
V = 2 x 10? ³

The formula for current is I = neAV.
Rearranging for n (number of free electrons per unit volume):
n = I / (eAV)
n = 10 / (1.6 * 10? ¹? * 5 * 10? * 2 * 10? ³)
n = 625 * 10²?

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A galvanometer of resistance G is converted into a voltmeter of range 0 - 1V by connecting a resistance R? in series with it. The additional resistance that should be connected in series with R? to increase the range of the voltmeter to 0 – 2V will b

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Four resistance 40Ω, 60Ω, 90Ω, and 110Ω make the arms of a quadrilateral ABCD. Across AC is a battery of emf 40 V and internal resistance negligible. The potential difference across BD in V is

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i? = 40/100 = 2/5
i? = 40/200 = 1/5
V? – V? = 40i? = 40 * 2/5
V? – V? = 16
V? – Vd = 90i? = 90/5 = 18
V? – Vd = 18 – 16 = 2 volt

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A circuit to verify Ohm's law uses ammeter and voltmeter in series or parallel connected correctly to the resistor. In the circuit:

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In Ohm's law experiment, ammeter is used in series because in series same current will flow through it. But voltmeter is used in parallel to resistor to measure the potential difference across it.

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The value of current i₁ flowing from A to C in the circuit diagram is:

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i? = 8/8 = 1 A

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In the figure shown, the current in the 10 V battery is close to:

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A battery of 3.0 V is connected to a resistor dissipating 0.5 W of power. If the terminal voltage of the battery is 2.5 V, the power dissipated within the internal resistance is:

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