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marishachu [46]

3 years ago

7

A circuit has a source voltage of 15V and two resistors in series with a total resistance of 4000Ω .If RI has a potential drop o

f 9.375 V across it, what is the value of R2? (a) 2000Ω (b) 1500Ω (c) 1000Ω (d) 500Ω

1 answer:

anastassius [24]3 years ago

6 0

Answer:

1500Ω

Explanation:

Given data

voltage = 15 V

total Resistance = 4000Ω

potential drop V = 9.375 V

To find out

R2

Solution

we know R1 +R2 = 4000Ω

So we use here Ohm's law to find out current I

current = voltage / total resistance

I = 15 / 4000 = 3.75 × $10^{-3}$ A

Now we apply Kirchhoffs Voltage Law for find out R2

R2 = ( 15 - V ) / current

R2 = ( 15 - 9.375 ) / 3.75 × $10^{-3}$

R2 = 1500Ω

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2 years ago

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saul85 [17]

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d. The alternate concentrated load is more critical to bending , shear and deflection

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The given parameters of the beam the beam are;

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b. The reduced floor live load, L in psf. is given as follows;

$L = L_o \times \left ( 0.25 + \dfrac{15}{\sqrt{k_{LL} \cdot A_T} } \right)$

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d. For the uniformly distributed load, we have;

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For the alternate concentrated load, we have;

$P_L$ = 1000 lb

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$V_{max}$ = 1,000 + 320 × 26/2 = 5,160 lbs

$M_{max}$ = 1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs

$v_{max}$ = 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI

Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load

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