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3 years ago

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A circuit contains a 40 ohm resistor and a 60 ohms resistor connected in parallel. If you test this circuit with a DMM you shoul

d read a total resistance

1 answer:

lorasvet [3.4K]3 years ago

4 0

<h2>Answer:</h2>

24Ω

<h2>Explanation:</h2>

When resistors are connected in parallel, the reciprocal of their combined resistance, when read with a DMM (Digital Multimeter - for measuring various properties of a circuit or circuit element such as resistance...) is the sum of the reciprocals of their individual resistances.

For example if two resistors of resistances R₁ and R₂ are connected together in parallel, the reciprocal of their combined resistance Rₓ is given by;

$\frac{1}{R_x}$ = $\frac{1}{R_1}$ + $\frac{1}{R_2}$

Solving for Rₓ gives;

$R_{x}$ = $\frac{R_1 * R_2}{R_1 + R_2}$ ------------------(i)

From the question;

Let

R₁ = resistance of first resistor = 40Ω

R₂ = resistance of second resistor = 60Ω

Now,

To get their combined or total resistance, Rₓ, substitute these values into equation (i) as follows;

$R_{x}$ = $\frac{40 * 60}{40 + 60}$

$R_{x}$ = $\frac{2400}{100}$

$R_{x}$ = 24 Ω

Therefore, the total resistance is 24Ω

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Given:

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Now, reactance $X_{L}$ at f' = 45.0 Hz:

$X'_{L} = 2\pi f'L$

$X'_{L} = 2\pi\times 45.0\times 0.13263 = 37.5\ohm$

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$I_{L(max) = \sqrt {\frac{2V_{rms}}{X'_{L}}}$

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3 years ago

I want to solve the question

DedPeter [7]

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Explanation:

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The maximum shear stress in the tube when the power is transmitted through a 4: 1 gearing is 28.98 MPa.

<h3>What is power?</h3>

Power is the energy transferred per unit time.

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The gear ratio Ne / Ns =4/1

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τ max = 16 x 190.96 x 0.038 /π x (0.038⁴ - 0.03⁴)

τ max = 28.98 MPa.

Thus, the maximum shear stress in the tube is 28.98 MPa.

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2 years ago

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3 years ago

An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no

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Answer:

Assumption:

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2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus $s_{1} =s_{2}$

From the refrigerant table A11-A13

$P_{1} =0.8MPa$ $\left \{ {{ {{v_{1}=v_{g} @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g} @0.8MPa =246.82 kJ/kg } - also {{s_{1}=s_{g} @0.8MPa =0.91853 kJ/kgK } } \right.$

sat vapor

m= $\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339} = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) = 38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}$

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b.) We take the content of the cylinder as the sysytem.

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$E_{in} - E_{out} =$ ΔE

$w_{b, out} =$ ΔU

$w_{b, out} =m([tex]u_{1} -u_{2$ )

$w_{b, out} =$ workdone during the isentropic process

=5.8491(246.82-219.9)

=5.8491(26.91)

=157.3993

=157.4kJ

4 0

3 years ago