Question

An alternating emf of 230V, 50Hz is connected across a pure ohmic resistance of 50Ω. Find (1) the current (2) equations for instantaneous values of current and voltage.

Answer 1

Given:

e_rms = 230 V, f = 50 Hz, R = 50 Ω  

1. R.M.S value of current, i_rms = `"e"_"rms"/"R" = 230/50` = 4.6 A
2. a. Peak value of current,
   `"i"_0 = "i"_"rms" xx sqrt2` 
   = `4.6 xx sqrt2`
   = 6.5 A
   ∴ Equation for instantaneous value of current,
   i = i₀ sin 2πft
   = 6.5 sin (2 × π × 50 × t)
   ∴ i = 6.5 sin 100 πt
   b. Peak value of voltage,
   e₀ = `"e"_"rms" xx sqrt2 = 230 xx sqrt2 = 325.27`V
   ∴ Equation for instantaneous value of voltage,
   e = e₀ sin 2πft 
   = 325.27 sin (2 × π × 50 × t)
   ∴ e = 325.27 sin 100 πt 
   i. R.M.S value of current is 4.6 A.
   ii. Equations for instantaneous value of current and voltage are i = 6.5 sin 100 πt and e = 325.27 sin 100 πt respectively.