Answer to Question #281865 in Electric Circuits for Ali Atwi

Question

An electric circuit is formed of :

_a generator delivering a constant voltage Ugenerator=24 volt

_Resistor D of resistance R=48 ohm

_Rheostat (D') of adjustable resistance R_h

1_Show applying law of addition of voltages that the current is given by the expression:

I=U_generator/R+R_h

2_We vary the value of R_h between two limiting values 0 ohm and 120 ohm.Calculate the value I₁ of I for R_h=0 ohm.Then calculate the value I₂ of I for R_h=120 ohm.

3_Deduce the role of a rheostat in a series connection.

4_Let P be the power dissipated by R.Give the expression of P in terms of R and I.Calculate the value P₁ of P for I=I₁ and the value of P₂ of P for I=I₂

5_The maximum power that the resistor D can stand is 5 watt. Show that resistor D can be damaged only in one of the two limiting values of R_h.

Answer 1

Let $I$ = the current in the circuit

Part 1

pd across $D= IR$

pd across $D'=IR_h$

$IR+IR_h= U_{generator}\\I= \frac{U_{generator}}{R+R_h}$

Part 2

$I_1=\frac{24}{48+0}=0.5A$

$I_2= \frac{24}{48+120}=0.1429A$

Part 3

It regulates the current flowing in a circuit

Part 4

$P= I^2R$

$P_1= 0.5^2×48=12 W$

$P_2=O.1429^2×48=0.9802W$

Part 5

When $R_h=0\Omega\>\>I_1=0.5A\>\>and$

$P=12W$

Since 12W>5W, the resistor D will be damaged by lower limit of $R_h$

When $R_h=120\Omega\,\,I_2=0.1429A$

and $P=0.9802W$

0.9802W<5W

The resistor D will not be damaged by upper limit of $R_h$

Therefore, the resistor D can be damaged only in one of the two limiting values of the $R_h$