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nikklg [1K]
3 years ago
11

A parallel circuit has two 8.0-ohm resistors and a power source of 9.0 volts. If a 12.5-ohm resistor is added to the circuit in

parallel, how will the current be affected and what value will it have?
Physics
2 answers:
Daniel [21]3 years ago
5 0

Answer:

1/R1 + 1/R2 + ... = 1/Re

So...

1/17.2 + 1/22.4 = 1/Re

0.1021 = 1/Re

Re = 9.792 Ohms

Now use the Voltage equation V = IR

6 = I * 9.792

I = 6/9.792 = 0.613 Amps.Pato 0.61

Explanation:

natta225 [31]3 years ago
4 0
The initial equivalent resistance of the circuit is
\frac{1}{R_{eq}}= \frac{1}{R_1}+ \frac{1}{R_2}= \frac{1}{8 \Omega}+ \frac{1}{8 \Omega} =  \frac{1}{4 \Omega}
which means 
R_{eq}= 4 \Omega
Therefore the initial current in the circuit is
I= \frac{V}{R}= \frac{9 V}{4 \Omega}=2.25 A

When the new resistor of 12.5 \Omega is added to the circuit in parallel, the new equivalent resistance of the circuit is
\frac{1}{R_{eq}} = \frac{1}{8 \Omega} + \frac{1}{8 \Omega}+ \frac{1}{12.5 \Omega}=  0.33 \Omega^{-1}
from which we find
R_{eq}=3 \Omega
This means that the equivalent resistance of the circuit has decreased, and the new current is
I= \frac{V}{R_{eq}}= \frac{9 V}{3 \Omega}=3 A
which means that the current in the circuit has increased.
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Radiation makes it impossible to stand close to a hot lava flow. Calculate the rate of heat transfer by radiation, in kW, from 1
Leya [2.2K]

Answer:

1.5 x 10⁵ W

Explanation:

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T = Temperature of fresh lava = 1000 °C = 1000 + 273 = 1273 K

T₀ = Temperature of surrounding = 25.3 °C = 25.3 + 273 = 298.3 K

ε = emissivity of the lava = 0.97

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Rate of  transfer is given as

E = σ ε A (T⁴ - T₀⁴)

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3 0
3 years ago
A 20.00-ohm, 5.00-watt resistor is placed in series with a power supply.(a) What is the maximum voltage that can be applied to t
Nutka1998 [239]
<h2>Answer:</h2>

(a) 10.00V

(b) 0.5V

<h2>Explanation:</h2>

(a) The power(P) supplied by a resistor of resistance (R) when a voltage(V) is passed across is given by;

P = V² / R       ------------------(i)

From the question;

P = 5.00W

R = 20.00Ω

Substitute these values into equation (i) as follows;

5.00 = V² / 20.00

V² = 5.00 x 20.00

V² = 100.00

V = \sqrt{100.00}

Solve for V;

V = 10.00V

Therefore, the maximum voltage that can be applied without harming the resistor is 10.00V

(b) By Ohm's law the current (I) flowing through a resistor of resistance (R) when a voltage (V) is applied is given by;

V = I x R

=> I = V / R             ----------------(ii)

Substitute the values of V and R into equation (ii) as follows;

I = 10.00 / 20.00

I = 0.5A

Therefore, the current through the resistor is 0.5A

7 0
3 years ago
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