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-BARSIC- [3]
3 years ago
11

A battery is connected to a resistor. Increasing the resistance of the resistor will __________. A battery is connected to a res

istor. Increasing the resistance of the resistor will __________. decrease the current in the circuit not affect the current in the circuit increase the current in the circuit
Engineering
1 answer:
belka [17]3 years ago
8 0

Answer: the increase in the external resistor will affect and decrease the current in the circuit.

Explanation: A battery has it own internal resistance, r, and given an external resistor of resistance, R, the equation of typical of Ohm's law giving the flow of current is

E = IR + Ir = I(R + r)........(1)

Where IR is the potential difference flowing in the external circuit and Or is the lost voltage due to internal resistance of battery. From (1)

I = E/(R + r)

As R increases, and E, r remain constant, the value (R + r) increases, hence the value of current, I, in the external circuit decreases.

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A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The air enters the
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Answer:

A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

We are given;

Temperature at state 1; T1 = 290 K

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A) The power assuming constant specific heats at room temperature is gotten from;

W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in

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T4 = [T3 × (r_p)^((1 - k)/k)]

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r_p = 760/95

r_p = 8

Also,k which is specific heat capacity of air has a constant value of 1.4

Thus;

Plugging in the relevant values, we have;

T4 = [(1100 × (8^((1 - 1.4)/1.4)]

T4 = 607.25 K

T2 = [290 × (8^((1.4 - 1)/1.4)]

T2 = 525.32 K

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W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000

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B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

At T3 = 1100 K, h3 = 1161.07 KJ/K

At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

Thus;

W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

W' = 17113.87 KW

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