# The voltage(V) volts in a circuit

1. Aug 21, 2010

### BonBon101

1. The problem statement, all variables and given/known data
The voltage(V) volts in a circuit at time(t) is given by

V(t)=18(1 - e -4t/9)

Find the expression for dV/dt.

Hence evaluate dV/dt at

(a)t=0 and (b)t=5 seconds

2. Relevant equations

Just wondering if this is right and if not where did i go wrong

3. The attempt at a solution

dV/dt=18(-4/9)(1-e-4/9t)

= -9(1-e-4/9t)

(a) t=0 -8(1-e-4/9(0))

=0mA

(b)t=5 -8(1-e-4/9(5))

=-7.13mA

2. Aug 21, 2010

### Feldoh

$$\frac{d}{dt}V(t) = \frac{d}{dt}18 - \frac{d}{dt}18e^{-4t/9}$$ what's the derivative?

3. Aug 21, 2010

### BonBon101

not to sure.Im sorry but im not that good at this hoping to get better!

4. Aug 22, 2010

### Feldoh

Well

$$\frac{d}{dt}V(t) = \frac{d}{dt}18 - \frac{d}{dt}18e^{-4t/9}$$

Whats the derivative of 18 with respect to t?

What about 18exp(-4t/9) with respect to t?

I seperated the parts of the equation using the fact that $$\frac{d}{dt}[v(t) + w(t)] = \frac{d}{dt}[v(t)] + \frac{d}{dt}[w(t)]$$ you can generally remember this rule as "the derivative of the sum is the sum of the derivatives".

Does this make sense?