Question

A current of 2.0 A is measured when a battery is connected across a resistor, R1 = 10Ω. When the same battery is connected across another resistor, R2 = 6.0Ω resistor, the current changes to 3.0 A. 

(a)          Calculate the potential differences across each of resistors.

(3 marks)

(b)          Find the emf and the internal resistance of the battery.

(9 marks)

(c) Determine the power dissipated in each resistor.

(3 marks) 

Answer 1

Solution.

$I_1=2.0A;$

$R_1=10\Omega;$

$I_2=3.0A;$

$R_2=6.0\Omega;$

$a) V_1=I_1R_1;$

$V_1=2.0\sdot10=20V;$

$V_2=3.0\sdot6.0=18V;$

$b) V=I(R+r);$

$V=I_1(R_1+r);$

$V=I_2(R_2+r);$

$I_1(R_1+r)=I_2(R_2+r);$

$I_1R_1+I_1r=I_2R_2+I_2r;$

$I_1r-I_2r=I_2R_2-I_1R_1$ ;

$r=\dfrac{I_2R_2-I_1R_1}{I_1-I_2};$

$r=\dfrac{3.0\sdot6.0-2.0\sdot10}{2.0-3.0}=2.0\Omega;$

$V=3.0(6.0+2.0)=24V;$

$c) P_1=I_1^2R_1;$

$P_1=2.0^2\sdot10=40.0W;$

$P_2=I_2^2R_2;$

$P_2=3.0^2\sdot6.0=54.0W;$

Answer: $a)V_1=20.0V;V_2=18.0V;$

$b)r=2.0\Omega; V=24.0V;$

$c)P_1=40.0W; P_2=54.0W.$