0.200A
0.070A
0.050A
0.008A
Correct answer is D
Given Data: L = 10, V = 50, F = 100, I = ?
Inductive Reactance [X\(_L\)] = 2\(\pi\)FL → 2 * 3.14 * 100 * 10
X\(_L\) = 6,280
Current[I] = \(\frac{V}{X_L}\) → \(\frac{50}{6280}\)
I = 0.0079
≈ 0.008A
2.5\(\Omega\)
5.0\(\Omega\)
8.0\(\Omega\)
10.0\(\Omega\)
Correct answer is B
r = resistance of the galvanometer
l = current through galvanometer = 20mA or 0.02A
V1 = p.d across the galvanometer = I X r = 0.02 X r
: V1 = 0.02r
V2 = p.d across the multiplier = 8 - 0.02r
R = resistance of the multiplier = 395Ω
where R = \(\frac{V}{I}\)
--> 395 = \(\frac{8−0.02r}{0.02}\)
CROSS MULTIPLY
8−0.02r = 395 * 0.02
8−0.02r = 7.9
--> 0.02r = 8 - 7.9
∴ r = \(\frac{0.1}{0.02}\)
= 5Ω
Which of the following statements about electric potential energy is not correct?
The electric potential energy of a positively charged particle increases when it moves to a point of higher potential
The electric potential energy of a negatively charged particle increases when it moves to a point of lower potential
The electric potential energy of a positively charged particle decreases when it moves to a point of higher potential
The work done in taking a charged particle around a closed path in an electric field is zero.
Correct answer is B
The potential energy for a positive charge increases when it moves against an electric field
And decreases when it moves with the electric field;
The opposite is true for a negative charge. Unless the unit charge crosses a changing magnetic field, its potential at any given point does not depend on the path taken.
0.90A
0.39A
0.30A
0.01A
Correct answer is C
Given Data: Emf = 1v, r = 2 ohms, R = 3 ohms, I = ?
3 resistance in parallel = \(\frac{1}{r_1}\) + \(\frac{1}{r_2}\) + \(\frac{1}{r_3}\)
\(\frac{1}{r_T}\)= \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\)
\(\frac{1}{r_T}\) = \(\frac{1+1+1}{2}\)
\(\frac{1}{r_T}\) = \(\frac{3}{2}\)
cross multiply
r\(_T\) = \(\frac{2}{3}\) or 0.67
E = I(R+r)
1 = I(3+0.67)
1 = I(3.67)
1 ÷ 3.67 = I
0.27 = I
: I ≈ 0.30A
maximum motor force if the current reverses its direction
no motor force of it is parallel to the field
no motor force if it is perpendicular to the field
a motor force with constant direction if either the current or the magnetic field is reversed
Correct answer is B
The greatest force is experienced when the conductor is at right angle to the field.
I.e at perpendicular or Ø = 90°