Please can you help out:

1)The Potential diff across the terminals of a battery is 8.5v when there is a current of 3amps in the cct from the -ve to the +ve terminal. When the current is 2amps in the reverse direction, the potential diff becomes 11volts.
a) what is the internal resistance of the battery?
b) what is the e.m.f of the battery

2) The instantaneous current I across a 200-ohm resistor is given by I= 10cos 377t
a) find the freq and the period
b) find the voltage across the resistor when t=1/240 second

At 3 amp,
8.5=emf =3r

at 2 amp
11 = emf=2r

you have two equations, two unknowns. Solve for emf and r. You can solve for rinternal directly by subtracting the equations.

On the second, frequency is in the given. remember Amplitude*cos(2PI*f*t) is the wave formula.

1) To solve this problem, we can use two key equations related to batteries and internal resistance.

a) Since the potential difference across the terminals of the battery is given by the equation V = emf - (current * internal resistance), we can substitute the given values to find the internal resistance (r). From the equation, we have:

8.5 = emf - (3 * r) (when the current is 3 amps)
11 = emf - (2 * r) (when the current is 2 amps)

To solve this system of equations, we can subtract the second equation from the first to eliminate emf:

(8.5 - 11) = (emf - (3 * r)) - (emf - (2 * r))
-2.5 = -r

Therefore, the internal resistance is 2.5 ohms.

b) To find the electromotive force (emf) of the battery, we can substitute the value of the internal resistance (r) into one of the equations above. Let's use the first equation:

8.5 = emf - (3 * 2.5)
8.5 = emf - 7.5
emf = 8.5 + 7.5
emf = 16 volts

Therefore, the electromotive force (emf) of the battery is 16 volts.

2) For the second problem, we are given an expression for the instantaneous current I across a 200-ohm resistor. Let's solve the two parts separately:

a) The given expression for current, I = 10cos(377t), tells us that the angular frequency (ω) is 377 rad/s. To find the frequency (f) and period (T), we can use the following relationships:

ω = 2πf (where ω is angular frequency and f is frequency)
T = 1/f (where T is period and f is frequency)

Substituting the value of ω, we have:

377 = 2πf
f = 377 / (2π)
f ≈ 60 Hz

To find the period T, we can substitute the value of frequency f into the second equation:

T = 1 / f
T = 1 / 60
T ≈ 0.0167 s

Therefore, the frequency is approximately 60 Hz and the period is approximately 0.0167 seconds.

b) To find the voltage across the resistor when t = 1/240 second, we can substitute this value into the expression for voltage:

Voltage = current * resistance

Since we know the current is given by I = 10cos(377t), and the resistance is 200 ohms, we can substitute the values and solve for the voltage. When t = 1/240 second:

Voltage = (10cos(377(1/240))) * 200
Voltage = (10cos(377/240)) * 200
Voltage ≈ (10 * 0.7854) * 200
Voltage ≈ 1570.8 volts

Therefore, the voltage across the resistor when t = 1/240 second is approximately 1570.8 volts.