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murzikaleks [220]

2 years ago

8

Sam constructs a circuit, connects a lead acid battery of 2 V to a lamp of resistance 3 Ω and places an ammeter across it. What

must be the reading of the ammeter?
A.
0.66 A
B.
0.5 A
C.
0.54 A
D.
0.61 A

2 answers:

Anon25 [30]2 years ago

8 0

It’s A because voltage equals current times resistance

frosja888 [35]2 years ago

4 0

Answer:

A. 0.66 Amps

Explanation:

Using ohms law, we can say that Voltage is equivalent to Current times Resistance. We are given the voltage and the resistance of the circuit, so we simply need to find the current.

V = IR

Solve for I, where V = 2volts and R = 3ohms.

V = IR

V * 1/R = I * R * 1/R

I = V/R

I = 2/3 Amps

Hence, we should choose option A, 0.66 Amps for the current in this simple circuit.

Cheers.

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2 years ago

A square loop of wire surrounds a solenoid. The side of the square is 0.1 m, while the radius of the solenoid is 0.025 m. The sq

Semmy [17]

Answer:

I=9.6×e^{-8} A

Explanation:

The magnetic field inside the solenoid.

B=I*500*muy0/0.3=2.1×e ^-3×I.

so the total flux go through the square loop.

B×π×r^2=I×2.1×e^-3π×0.025^2

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6 0

2 years ago

A meter stick can be read to the nearest millimeter and a travelling microscope can be read to the nearest 0.1 mm. Suppose you w

german

Answer: No

Explanation:

Length= 2cm= 20mm

Now meter stick can read to nearest millimeter.

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3 0

3 years ago

A shunt regulator utilizing a zener diode with an incremental resistance of 8 ohm is fed through an 82-Ohm resistor. If the raw

spayn [35]

Answer:

$\triangle V_0=0.08V$

Explanation:

From the question we are told that:

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Resistor Feed $R_f=82ohms$

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Generally the equation for voltage rate of change is mathematically given by

$\frac{dV_0}{dV}=\frca{R}{R_1r_3}$

Therefore

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7 0

2 years ago

The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet

alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

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$T_{c_i}=100^{\circ}C$

$T_{c_o}=125^{\circ}C$

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

$m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]$

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we can see that heat capacity of hot fluid is minimum

Also from energy balance

$Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )$

$NTU=\frac{UA}{\left ( mc_p\right )_{h}}$ = $\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}$

$T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}$

$T_m=41.63^{\circ}C$

NTU=1.921

$And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}$

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5 0

3 years ago