Two resistors of 4.0 Ω and 6.0 Ω are connected (a) in series, (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case draw a circuit diagram and calculate the current through the battery.
Current Electricity
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Question

Two resistors of 4.0 Ω and 6.0 Ω are connected (a) in series, (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case draw a circuit diagram and calculate the current through the battery.

Current Electricity

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KnowledgeBoat · Accepted Answer

(a) Circuit diagram showing two resistors of 4.0 Ω and 6.0 Ω connected in series with a battery of 6.0 V and negligible internal resistance is shown below: Given, Two resistors of 4 Ω and 6 Ω are connected in series. If the equivalent resistance of this part is R'_s then R'_s = (4 + 6) Ω = 10 Ω Potential Difference V = 6 V Current I = ? From Ohm's law V = IR Substituting the values in the formula above, we get, 6 = I x 10 I = 6 / 10 = 0.6 A Hence, in series, current through the battery = 0.6 A (b) Circuit diagram showing two resistors of 4.0 Ω and 6.0 Ω connected in parallel with a battery of 6.0 V and negligible internal resistance is shown below: Given, Two resistors of 4 Ω and 6 Ω are connected in parallel. If the equivalent resistance of this part is R_p then Potential Difference V = 6 V Current I = ? From Ohm's law V = IR Substituting the values in the formula above, we get, 6 = I x 2.4 I = 6 / 2.4 = 2.5 A Hence, in parallel, current through the battery = 2.5 A

Physics

Two resistors of 4.0 Ω and 6.0 Ω are connected (a) in series, (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case draw a circuit diagram and calculate the current through the battery.

Current Electricity

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Two resistors of 4.0 Ω and 6.0 Ω are connected (a) in series, (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case draw a circuit diagram and calculate the current through the battery.

Current Electricity

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