A resistors of 6 - is connected in series with another resistor of 4 - A potential difference of 20 V is applied across the combination-

Given : #160; #160; V = 20 volts #160; #160; #160; #160; #160; R_1 = 6Ω #160; #160; #160; #160; #160; #160; ...

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Byju's Answer

Standard XII

Physics

Parallel and Series Combination of Capacitors

A resistors o...

Question

A resistors of $6 Ω$ is connected in series with another resistor of $4 Ω$ . A potential difference of 20 V is applied across the combination.

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Solution

Given : $V = 20$ volts $R_{1} = 6 Ω$ $R_{2} = 4 Ω$
Total resistance of the circuit $R_{e q} = R_{1} + R_{2} = 6 + 4 = 10 Ω$
$∴$ Current flowing in the circuit $I = \frac{V}{R_{e q}} = \frac{20}{10} = 2$ A

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A resistor of $6 Ω$ is connected in series with another resistor of $4 Ω$ . A potential difference of 20 V is applied across the combination. Calculate:

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Two resistors of resistance 3 $Ω$ and 6 $Ω$ are in parallel and the combination is in series with a 4 $Ω$ resistor. A potential difference of 90V is applied across the network. The potential difference across 4 $Ω$ the resistor and the current in the 3 $Ω$ resistor are respectively,

Q. A battery of internal resistance,

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Two resistors 4 Ω and 2 Ω are connected in series, and potential difference of 12 V is applied across the combination. Then, the current through the circuit is:

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A resistors of 6Ω is connected in series with another resistor of 4Ω. A potential difference of 20 V is applied across the combination.

Given : V=20 volts R1=6Ω R2=4ΩTotal resistance of the circuit Req=R1+R2=6+4=10Ω∴ Current flowing in the circuit I=VReq=2010=2 A

A resistors of $6 Ω$ is connected in series with another resistor of $4 Ω$ . A potential difference of 20 V is applied across the combination.

Given : $V = 20$ volts $R_{1} = 6 Ω$ $R_{2} = 4 Ω$
Total resistance of the circuit $R_{e q} = R_{1} + R_{2} = 6 + 4 = 10 Ω$
$∴$ Current flowing in the circuit $I = \frac{V}{R_{e q}} = \frac{20}{10} = 2$ A