1. What are the maximum voltage and current which should occur a. for a (1/4) wa

Question

1. What are the maximum voltage and current which should occur a. for a (1/4) watt, 300ohms resistor? b. for a (1/4) watt, 3000 ohm resistor?

2.Answer Voltmeter or Ammeter
   a. has a low resitance
   b. has a high resistance
   c. is connected in parallel with a circuit element or elements
   d. the circuit must be rewired in order to include it
e. it can be added to the circuit without rewiring the circuit

3. For a given power dissipation, does V(max) increase or decrese with increasing R? Does I(max) increase or decrease with increasing R?

4. Given V=10 in series with R1=100ohms and R2=400ohms.

a. Calculate I=V/(R1+R2)

b. Calculate V1 and V2

c. Calculate the power dissipated in each resistor.

5. What is the expected slope of the plot in which the current (I) is along the horizontal axis and the voltage (V) is plotted on the veltical axis? What is the expected slope of the plot in which the current is along vertical axis and the voltage is plotted on the horizontal axis?

6. In an AC circuit with Vs=10V and series resistances R1=100ohms and R2=250ohms, find the rms current (I), rms voltages V1 and V2 and peak voltages V1p and V2p. Formula: V=Vpcos(2pift) or V=Vpsin(2pift) Vrms=Vp/SQRT2

Explanation / Answer

1. a)

Power = 1/4 W
Power = V x I, V and I are voltage and current respectively.
VI = I2R, (by substituting V = IR)
I2R = 1/4 W
I = sqrt[1/1200] = 0.02886 A.
V =IR = 8.66 V

b)

Here also, I2R = 1/4 W, but R = 3000 ohm
I = sqrt[1/12000] = 0.00912 A.
V = IR = 27.38 V.

2

a) Ammeter
It should have very small resistance so that all the current should pass through it for measurement.

b) Voltmeter
It should not draw current from the circuit, other wise it will affect the proper working of the circuit.

c) Voltmeter
Connections are made parallel to read voltage.

d) Ammeter.
Since ammeter is connected in series, it needs to be inserted in a circuit.

e) Voltmeter
Same explanation as above.

3)
On increasing the resistance current will decrease and hence the voltage will increase.

4. a)

Current I = V/(R1+R2) = 10/500 = 0.02 A.

b)

V1 = I x R1 = 0.02 x 100 = 2 V.
V2 = I x R2 = 0.02 x 400 = 8 V.

c)

Power in R1 = I x V1 = 0.02 x 2 = 0.04 W.
Power in R2 = I x V2 = 0.02 x 8 = 0.16 W.

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