Physics Questions and Answers

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

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.

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

The greatest force is experienced when the conductor is at right angle to the field.

I.e at perpendicular or Ø = 90°

The photoelectric effect is the emission of electrons when electromagnetic radiation, such as light, hits a material