Two resistors of $2.0 Ω$ and $3.0 Ω$ are connected (a) in series, (b) in parallel, with battery of 6.0 V and negligible internal resistance. For each case

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Solution

Two resistors of $2 Ω$ & $3 Ω$

(i) Reg. = $5 Ω$

I = $\frac{6}{5} A$

(ii) Reg. = $3 ∥ 2$ = $\frac{6}{5}$

$= \frac{6}{6} \times 5$

= 5 A

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Two resistors of $2.0 Ω$ and $3.0 Ω$ 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.

Two resistors of

2.0 Ω

and

3.0 Ω

are connected in parallel, with a battery of 6.0 V and negligible internal resistance.

The current in the circuit is :

Q. Three resistors of resistances

3 Ω, 4 Ω

and

5 Ω

are combined in parallel. This combination is connected to a battery of emf

12

V and negligible internal resistance, current through each resistor in ampere is?

Q. a) Three resistors of resistances

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and

4 Ω

are combined in series. What is the resistance of the combination?
b) If this combination is connected to a battery of emf 10 V and negligible internal resistance, obtain the potential drop across each resistor.

Q. Four resistors of resistances

0.5 Ω, 1.5 Ω, 4 Ω

and

6 Ω

are connected in series to a battery of EMF

6 V

and negligible internal resistance. Calculate the:
(a) current drawn from the cell.
(b) potential difference at the end of each resistor.
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Two resistors of 2.0Ω and 3.0Ω are connected (a) in series, (b) in parallel, with battery of 6.0 V and negligible internal resistance. For each case

Two resistors of 2Ω & 3Ω (i) Reg. = 5Ω I = 65A (ii) Reg. = 3∥2=65 =66×5 = 5 A

Two resistors of $2.0 Ω$ and $3.0 Ω$ are connected (a) in series, (b) in parallel, with battery of 6.0 V and negligible internal resistance. For each case

Two resistors of $2 Ω$ & $3 Ω$

(i) Reg. = $5 Ω$

I = $\frac{6}{5} A$

(ii) Reg. = $3 ∥ 2$ = $\frac{6}{5}$

$= \frac{6}{6} \times 5$

= 5 A