Answer
$I=9.60\times10^{-4}A$
$V=4.8V$
$E_I=-20%$
$E_V=20%$
Work Step by Step
Net resistance of circuit is
$R_{eq}=1.0\Omega+0.50\Omega+7.5k\Omega+(\frac{1}{15k\Omega}+\frac{1}{7.5k\Omega})^{-1}=12.5k\Omega$
The ammeter will read:
$I=\frac{V}{R_{eq}}=\frac{12.0V}{12.5k\Omega}=9.60\times10^{-4}A$
The voltmeter will read:
$R=(\frac{1}{15k\Omega}+\frac{1}{7.5k\Omega})^{-1}=5k\Omega$
$V=IR=(9.60\times10^{-4}A)(5k\Omega)=4.8V$
Without the voltmeter or ammeter, the net resistance of the circuit will be:
$R_{eq}=1.0\Omega+0.50\Omega+2\times7.5k\Omega=15k\Omega$
and current will be:
$I=\frac{V}{R_{eq}}=\frac{12.0V}{15k\Omega}=8.00\times10^{-4}A$
and voltage will be:
$V=IR=(8.00\times10^{-4}A)(7.5k\Omega)=6.0V$
Percent error in current measure is:
$E_I=\frac{8.00\times10^{-4}A-9.60\times10^{-4}A}{8.00\times10^{-4}A}\times100\%=-20%$
Percent error in voltage measure is:
$E_V=\frac{6.0V-4.8V}{6.0V}\times100\%=20%$
