# Self and mutual inductance of ideal autotransformer

Discussion in 'Homework Help' started by cjacobs, Mar 1, 2012.

1. ### cjacobs Thread Starter New Member

Dec 8, 2010
10
0
I am stuck on an assignment question can someone shed some light.

I need to determine V1 and V2 of an ideal autotransformer in terms of their self and mutual voltages induced across the 2 coils.

When I1 is flowing down through coil 1, can I2 flow up through coil 1 and assume that the mutual inductance in coil 1(L1) is (L1L2) (I1+I2) or do i need to subtract (I2-I1) because the primary I1 is smaller than the secondary I2.

Any help will be appreciated.

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Presumably you are considering a step down auto-transformer. L1 connects between the input and the output terminals. L2 connects from the output to the common terminal. L1 & L2 are ideally mutually coupled. Again one further assumes there is zero resistance in the inductive elements.

If one further states that the input voltage is Vin with current I1 and the output voltage is Vout with load current I2. Current I1-I2 flows in the L2 branch.

Vin=jωL1*I1+jωM*(I1-I2)+jωL2*(I1-I2)+jωM*I1
Vout=jωL2*(I1-I2)+jωM*I1

M=√(L1*L2)

These can then be further simplified.

3. ### cjacobs Thread Starter New Member

Dec 8, 2010
10
0
Thank You TNK for you response. I got the same two formulas after posting my thread. So the first part of the question is done and because there is no load on secondary I2=0, so i can simplify to get V1/V2 ratio.

Thanks Again.