Question #281865

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 Rh


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

I=Ugenerator/R+Rh


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




1
Expert's answer
2021-12-22T14:14:08-0500

Let II = the current in the circuit

Part 1


pd across D=IRD= IR

pd across D=IRhD'=IR_h

IR+IRh=UgeneratorI=UgeneratorR+RhIR+IR_h= U_{generator}\\I= \frac{U_{generator}}{R+R_h}

Part 2


I1=2448+0=0.5AI_1=\frac{24}{48+0}=0.5A


I2=2448+120=0.1429AI_2= \frac{24}{48+120}=0.1429A


Part 3


It regulates the current flowing in a circuit


Part 4


P=I2RP= I^2R


P1=0.52×48=12WP_1= 0.5^2×48=12 W


P2=O.14292×48=0.9802WP_2=O.1429^2×48=0.9802W


Part 5


When Rh=0ΩI1=0.5AandR_h=0\Omega\>\>I_1=0.5A\>\>and

P=12WP=12W


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


When Rh=120ΩI2=0.1429AR_h=120\Omega\,\,I_2=0.1429A

and P=0.9802WP=0.9802W


0.9802W<5W

The resistor D will not be damaged by upper limit of RhR_h

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












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