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viva [34]
2 years ago
10

A 20 Ohm resistance is connected

Physics
1 answer:
frutty [35]2 years ago
8 0

Answer: 225 V

Explanation:

<u>Given:</u>

Secondary voltage,V2 = 150v

Resistance is connected across secondary winding,∴R2 = 20Ω

Supply current, ie, I1 = 5A

\text{Now, secondary current,} \ $I_{2}=\frac{V_{2}}{R_{2}}=\frac{150}{20}=7.5$$$\therefore I_{2}=7.5 \mathrm{~A}$$For Transformers, we have $\frac{N_{2}}{N_{1}}=\frac{V_{2}}{V_{1}}=\frac{I_{1}}{I_{2}}$ so, Taking $\frac{N_{2}}{N_{1}}=\frac{l_{1}}{l_{2}}$ inserting all given \& obtained values, we get $\frac{N_{2}}{N_{1}}=\frac{5}{7.5}$$$\therefore \text { Turns ratio }=\frac{N_{1}}{N_{2}}=\frac{7.5}{5}=\frac{3}{2}=3: 2$$

\text{Again taking,}\frac{N_{1}}{N_{2}}=\frac{V_{2}}{V_{1}}$ \\So, \\$V_{1}=\frac{N_{1}}{N_{2}} V_{2}\\$ inserting all values, we get, $V_{1}=\frac{7.5}{5} * 150=225$\\Therefore, \fbox{Primary voltage, {$V_{1}=225 \mathrm{~V}$}}

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What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Yo
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Full Question

Consider two point charges located on the x axis: one charge, Q1 = -12.0 nC , is located at x1 = -1.705m ; the second charge, Q2 = 36.5 nC, is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Your answer may be positive or negative, depending on the direction of the force.

Answer:

6.79E6 N

Explanation:

Given

Q1 is negative and to the left of Q3 the force will be to the left

Q2 is positive and to the right of Q3 the the force will also be to the left

Net Force is calculated as:

Using Coulomb's law

Coulomb's law: F = kqQ / r²

the constant k = 8.99 x 10^9 N m2 / C2

F = -kQ1*Q3/(r1)² -kQ2*Q3/(r2)²)

F = -kQ3(Q1/(r1)² + Q2/(r2)²)

Where

Q1 = -12nC = -12 * 10^-9C

Q2 = 36.5nC = 36.5 * 10^-9C

Q3 = 49.5nC = 49.5 * 10^-9C

x1 = -1.705m

x2 = x = 0

x3 = -1.170m

r1 = x3 - x1

r1 = -1.170 - -1.705

r1 = -1.170 + 1.705

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F = (-8.99 * 10^9)(49.5 *10^-9) [-12 * 10^-9/0.286225 + 36.5 * 10^-9/1.3689]

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F = -445.005 * −1.5261314344648E−8

F = -(8.99 * 10^9) * (49.5 * 10^-9) * [ (-12 * 10^-9) /(-1.770 - -1.705)² + (36.5 * 10^-9)/(-1.170)²]

F = -445.005( −0.000002813572941777538)

F = 0.00000679136118994008324

F = 6.79E6 N

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