Answer
(a) $86\Omega$
(b) $1.76V, 8.48V, 13.76V$
(c) greater
Work Step by Step
(a) We know that
$R=R_{eq}-11\Omega-53\Omega$....eq(1)
Now $R_{eq}=\frac{V}{I}$
We plug in the known values to obtain:
$\implies R_{eq}=\frac{24.0V}{0.16A}$
$\implies R_{eq}=150\Omega$
We plug in this value in eq(1), to obtain:
$R_{eq}=150\Omega-11\Omega-53\Omega$
$\implies R_{eq}=86\Omega$
(b) We can determine the required potential difference as follows:
$V_{11}=IR_{11}$
$\implies V_{11}=11\Omega\times 0.16A$
$\implies V_{11}=1.76V$
similarly $V_{53}=53\Omega\times 0.16A$
$V_{53}=8.48V$
and $V_{86}=86\Omega\times 0.16A$
$\implies V_{86}=13.76V$
(c) According to Ohm's law $R_{eq}=\frac{V}{I}$. Thus, when the value of V is greater than 24V, then the equivalent resistance should be more to get a constant current. Hence, the resistance $R$ is greater than that determined in part (a) -- that is, $86\Omega$.
