In the given circuit, the voltage across load resistance $\left(R _L\right)$ is:
(1) $8.75 V$
(2) $9.00 V$
(3) $8.50 V$
(4) $14.00 V$
$i=\frac{14}{4}=3.5 mA$
$V _L=iR _L=3.5 \times 2.5 volt$
$=8.75$ volt
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