Ask question

Ask question

All categories

VashaNatasha [74]

2 years ago

7

An electric heater containing two heating wires X and Y is connected to a power supply of electromotive force(emf) 9.0V and negl

igible internal resistance as shown in fig 6.1
Wire X has a resistance of 2.4 ohm and wire Y has a resistance of 1.2ohm. A voltmeter is connected in parallel with the wires. A variable resistor is used to adjust the power dissipated in wires X and Y.

The variable resistor is adjusted so that the voltmeter reads 6.0V.

Calculate the resistance of the variable resistor.

1 answer:

mina [271]2 years ago

7 0

Answer:

0.4 ohms.

Explanation:

From the circuit,

The voltage reading in the voltmeter = voltage drop across each of the parallel resistance.

1/R' = 1/R1+1/R2

R' = (R1×R2)/(R1+R2)

R' = (2.4×1.2)/(2.4+1.2)

R' = 2.88/3.6

R' = 0.8 ohms.

Hence the current flowing through the circuit is

I = V'/R'................ Equation 1

Where V' = voltmeter reading

I = 6/0.8

I = 7.5 A

This is the same current that flows through the variable resistor.

Voltage drop across the variable resistor = 9-6 = 3 V

Therefore, the resistance of the variable resistor = 3/7.5

Resistance = 0.4 ohms.

You might be interested in

Recall that in the equilibrium position, the upward force of the spring balances the force of gravity on the weight. Use this co

natima [27]

Recall that in the equilibrium position, the upward force of the spring balances the force of gravity on the weight is given below.

Explanation:

Measure unstretched length of spring, L. E.g. L = 0.60m.

Set mass to a convenient value (e.g. m = 0.5kg).

Hang mass.

Measure new spring length, L'. E.g. L' = 0.70m.

Calculate extension: e = L' - L = 0.70 – 0.60 = 0.10m

Use mg = ke (in equilibrium weight = tension)

k = mg/e

Don't know what value you are using for example. Suppose it is 10N/kg (same thing as 10m/s²).

k = 0.5*10/0.10 = 50 N/m

Repeat for a few different masses. (L always stays the same.)

Take the average of your k values.

5 0

1 year ago

Read 2 more answers

Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + 1 4 u' + 2

GarryVolchara [31]

Answer:

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

Explanation:

Given vibrating system is

$u''+\frac{1}{4}u'+2u= 2cos \omega t$

Consider U(t) = A cosωt + B sinωt

Differentiating with respect to t

U'(t)= - A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U''(t) = - A ω² cosωt -B ω² sin ωt

Putting this in given equation

$-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t$

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t$

Equating the coefficient of sinωt and cos ωt

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2$

$\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0$ .........(1)

and

$\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0$

$\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0$ ........(2)

Solving equation (1) and (2) by cross multiplication method

$\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}$

$\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}$

$\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}$ and $B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}$

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

5 0

2 years ago

An apple is whirled round in a horizontal circle on the end of a string which is tied to the stalk. It is whirled faster and fas

Brrunno [24]

when the apple moves in a horizontal circle, the tension force in the string provides the necessary centripetal force to move in circle. the tension in the string is given as

T=mv²/r

where T = tension force in the string , m = mass of the apple

v = speed of apple , r = radius of circle.

clearly , tension force depends on the square of the speed. hence greater the speed, greater will be the tension force.

at some point , the speed becomes large enough that it makes the tension force in the string becomes greater than the tensile strength of the string. at that point , the string breaks

6 0

2 years ago

Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of

schepotkina [342]

If speed = distance/time , then time = speed/distance.

So...

Speed of light = 3*10^8(m/s)
Average distance from Earth to Sun = 149.6*10^9(m)

Therefore, t=(3*10^8(m/s))/(149.6*10^9(m))

I hope this was a helpful explanation, please reply if you have further questions about the problem.

Good luck!

5 0

1 year ago

A cylindrical rod of steel (E = 207 GPa, 30 × 10 6 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a

Yanka [14]

Answer:

Diameter of the cylinder will be $d=2.998\times 10^4m$

Explanation:

We have given young's modulus of steel $E=207GPa=207\times 10^9Pa$

Change in length $\Delta l=0.38mm$

Length of rod $l=500mm$

Load F = 11100 KN

Strain is given by $strain=\frac{\Delta l}{l}=\frac{0.38}{500}=7.6\times 10^{-4}$

We know that young's modulus $E=\frac{stress}{strain}$

So $207\times 10^9=\frac{stress}{7.6\times 10^{-4}}$

$stress=1573.2\times 10^{-5}N/m^2$

We know that stress $=\frac{force}{artea }$

So $1573.2\times 10^{-5}=\frac{11100\times 1000}{area}$

$area=7.055\times 10^{8}m^2$

So $\frac{\pi }{4}d^2=7.055\times 10^{8}$

$d=2.998\times 10^4m$

6 0

2 years ago