Question

a battery of emf 12 V and internal resistance 2 ohm is connected with two resistors A and B of resistance 4 ohm and 6 ohm respectively joined in series. find 1) current in the circuit. 2) terminal voltage of the cell. 3) potential difference across 6 ohm resistor. 4) electrical energy spent per minute in 4 ohm resistor. ​

Answer 1

Answer: 1.) 1A

2.) 10V

3.) 6V

4.) 240J

Explanation: 1.) i=V/(R+r) = 12/(10+2)=12/12=1A

2.) Terminal voltage= E-ir

V=12-(1*2) =12-2 = 10V

3.) V=iR = 1A*6ohm = 6V

4.) Energy= i²Rt = 1² *4*60 = 240J

Answer 2

(A) Current will be 1 amp

(B) Terminal voltage will be 10 volt

(C) Potential difference across 6 ohm resistor will be 6 volt

(D) Energy spent per minute in 4 ohm resistor will be 240 J

Explanation:

We have given emf of the battery E = 12 volt

Internal resistance r = 2 ohm

It is given two resistance A and B of resistances 4 ohm and 6 ohm are connected in series

So equivalent external resistance R = 4+6 = 10 ohm

Total resistance = 10+2 = 12 ohm

(A) According to ohm's law V = IR

So current $i=\frac{E}{R}=\frac{12}{12}=1amp$

(B) We know that V = E - ir

So $V=12-1\times 2=12-2=10volt$

So terminal voltage will be 10 volt

(C) Potential difference across 6 ohm resistor

$V=iR=1\times 6=6volt$

(D) 1 minute = 60 sec

So energy spent in 4 ohm resistor $E=i^2Rt=1^2\times 4\times 60=240J$

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