In an NPN transistor circuit, the collector current is 10 mA. If 90% of the electrons emitted reach the collector, the emitter current (${\mathrm{i}}_{\mathrm{E}}$) and base current (${\mathrm{i}}_{\mathrm{B}}$) are given by
(a) ${\mathrm{i}}_{\mathrm{E}}$ = –1 mA, ${\mathrm{i}}_{\mathrm{B}}$ = 9 mA
(b) ${\mathrm{i}}_{\mathrm{E}}$ = 9 mA, ${\mathrm{i}}_{\mathrm{B}}$ = – 1 mA
(c) ${\mathrm{i}}_{\mathrm{E}}$ = 1 mA, ${\mathrm{i}}_{\mathrm{B}}$ = 11 mA
(d) ${\mathrm{i}}_{\mathrm{E}}$ = 11 mA, ${\mathrm{i}}_{\mathrm{B}}$ = 1 mA

(d)

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