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tankabanditka [31]
2 years ago
7

You have two resistors with the same cross sectional area and resistivity. Resistor A has length L1 and resistor B has length L2

. Assume L2 > L1. If the resistors are connected in series with a battery, which resistor has the most power delivered to it? Explain. b: If they are connected in parallel with a battery, which resistor has the most power delivered to it? Explain.
Physics
1 answer:
oee [108]2 years ago
8 0

Answer:

Explanation:

Given

Resistor A has length L_1

and Resistor B has Length L_2

and Resistance is given by

R=\frac{\rho L}{A}

Considering \rhoand A to be constant thus

R_2>R_1 because L_2>L_1

(a)When they are connected in series

As the current in series is same and power is i^2R

therefore P_2>P_1 as R is greater for second resistor

(b)if they are connected in Parallel

In Parallel connection Voltage is same

P=\frac{V^2}{R}

resistance of 2 is greater than 1 thus Power delivered by 1 is greater than 2

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Now

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Total time:-

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The light bulb converts electrical energy into light and ____. A) chemical B) electromagnetic C) heat D) nuclear
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7 0
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Read 2 more answers
the ratio of the energy per second radiated by the filament of a lamp at 250k to that radiated at 2000k, assuming the filament i
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Answer:

(a) \frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}

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(a)

The power radiated from a black body is given by Stefan Boltzman Law:

P = \sigma AT^4

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P = Energy Radiated per Second = ?

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So the ratio of power at 250 K to the power at 2000 K is given as:

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(b)

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3 years ago
How the Law of Conservation of Matter is supported by the experimental demonstrations?
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