Hi,Was wondering if anyone could help me with this question?

The Voltage (Vs) is 2.5volts A.C and has an output impedance of 1 ohm and forms a potential divider with a lamp,which has a resistance of 12.5 ohms.

1)What is the % error caused by connecting an analogue meter across the lamp?
2)What is the systematic error due to the instruments spec?

The input impedance for meter is 9kohms per volt
The allowance is plus or minus 4% of fs
The scale is 0-10V.

Thanks

First convert the voltage source of 2.5V in series with a 1 ohm resistor using a Thevinin/Norton equivalent and then find the voltage across the lamp. Call that Vlamp.

once you know Vlamp and you can compute the Z of the meter. 9K ohms/volt * Vlamp - Rmeter gives answer in ohms.

Since V and R are linear, you can get the % difference of RLamp and the parallel combination of Rlamp and Rmeter. That's the answer to #1. This error will have a sign. The meter will read lower, so it's negative %.